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THE  ELEMENTS   OF  THE 
DIFFERENTIAL  AND  INTEGRAL  CALCULUS 


?&&&' 


THE   ELEMENTS 


OF  THE 


DIFFERENTIAL  AND  INTEGRAL 


CALCULUS 


WITH  NUMEROUS  EXAMPLES 


BY 
DONALD   FRANCIS   CAMPBELL,  Ph.D. 

PROFESSOR  OF  MATHEMATICS,  ARMOUR  INSTITUTE  OF  TECHNOLOGY 


BOSTON  COLLEGE  LIBRARY 

CHEST'  ILL,  MASS. 


**AT(i-  oePTt 


THE    MACMILLAN   COMPANY 

LONDON :  MACMILLAN  &  CO.,  Ltd. 
1904 

All  rights  reserved 


Copyright,  1904, 
By  THE  MACMILLAN  COMPANY. 


Set  up  and  electro  typed.     Published  October,  1904. 


Nottooot)  $ress 

J.  S.  Cushing  &  Co.  — Iierwick  &  Smith  Co. 

Norwood,  Mass.,  U.S.A. 


PREFACE 

This  book  was  written  to  meet  the  needs  of  my  own  classes ; 
yet  it  is  hoped  that  not  only  teachers  of  mathematics  in  tech- 
nical colleges,  but  those  in  classical  colleges  and  universities 
as  well,  will  find  it  suitable  for  a  first  course  in  the  Dif- 
ferential and  Integral  Calculus. 

In  many  technical  colleges,  among  them  the  one  with  which 
I  am  connected,  the  study  of  Calculus  is  begun  in  the  first 
year  of  the  course.  As  such  an  arrangement  involves  begin- 
ning a  difficult  branch  of  mathematics  with  somewhat  imma- 
ture students,  the  first  few  chapters  in  both  the  Differential 
and  Integral  parts  are  discussed  in  more  detail  than  is  usual 
in  text-books. 

Throughout  the  book  I  have  confined  myself  strictly  to 
those  subjects  which  I  know  from  my  own  experience  are 
most  needed  by  my  own  students.  It  seemed  wise  to  me 
to  omit  all  subjects  only  remotely  connected  with  those  of 
engineering,  and  introduce  a  few  elementary  chapters  in 
Mechanics.  Thus  I  was  able,  without  encumbering  the  book, 
to  afford  a  short  introduction  to  Mechanics  and  Differential 
Equations  as  well  as  to  view  the  principles  of  Attraction, 
Centers  of  Gravity,  and,  to  a  certain  extent,  Moments  of 
Inertia,  from  the  mechanical  rather  than  the  purely  mathe- 
matical side.  If  the  teacher  feels  that  he  should  treat  any 
subject  omitted  here,  he  can  readily  do  so  by  lecture. 

The  part  of  the  book  which  differs  most  widely  from  other 
books  is  that  dealing  with  the  Integral  Calculus.  It  has  been 
my  experience  that  a  working  knowledge  of  the  principles  of 
the  Integral  Calculus  can  be  gained  only  by  a  careful  con- 


vi  PREFACE 

sideration  of  the  details  of  the  subject.  I  have,  therefore, 
even  at  the  risk  of  being  considered  prolix,  entered  into  a  full 
explanation  of  each  step  in  the  formation  of  each  summation 
and  integral. 

It  was  my  good  fortune  to  begin  the  study  of  Mechanics 
and  the  Differential  and  Integral  Calculus  with  MacGregor's 
Kinematics  and  Dynamics  and  Byerly's  works  in  Calculus, 
under  the  respective  authors.  To  these  books  and  men  I  am 
indebted  for  help  probably  more  than  I  am  aware  of  being. 
I  have  not  consciously  followed  any  of  their  methods  of 
treatment,  however,  although  I  frequently  consulted  the  work 
in  Kinematics  and  Dynamics  during  the  preparation  of  the 
first  two  chapters  in  Mechanics.  To  Appel's  Course  d' 'Ana- 
lyse I  am  indebted  for  the  proof  given  in  Art.  186  as  well  as 
for  other  valuable  hints.  To  B.  0.  Peirce's  Table  of  Inte- 
grals, Glanville  Taylor's  Calculus,  Gibson's  Calculus,  Tait  and 
Steele's  Dynamics  of  a  Particle,  and  Ziwet's  Mechanics  I  also 
made  frequent  reference.  A  few  problems  I  have  taken  from 
Harvard  examination  papers. 

To  a  colleague,  Professor  N.  C.  Piggs,  I  am  under  obliga- 
tions for  an  extremely  careful  revision  of  both  the  manuscript 
and  proofs,  for  a  number  of  problems  with  answers,  and  for 
verification  of  a  large  number  of  my  answers.  To  another 
colleague,  Professor  C.  W.  Leigh,  I  am  indebted  for  verifica- 
tion of  a  number  of  answers.  To  a  third-year  student  of  the 
Armour  Institute  of  Technology,  Mr.  Garfield  Lennartz,  I  am 
indebted  for  whatever  of   mechanical  excellence  the  figures 

possess. 

D.  F.  CAMPBELL. 

Chicago, 

August  8, 1904. 


CONTENTS 


PAGES 

Introduction .     1-9 

CHAPTER   I 
Limits 10-29 

CHAPTER   II 
Derivatives 30-34 

CHAPTER   III 

Geometrical  Interpretation  of  a  Derivative.       Problems  in 

Speed 35-46 

CHAPTER   IV 
General  Formulas  for  Differentiation     ....         47-58 

CHAPTER  V 
Successive  Differentiation 59,  60 

CHAPTER   VI 
Applications  of  Derivatives  to  Curves.     Maxima  and  Minima  61-78 

CHAPTER   VII 

Formulas  for  Differentiation  of  Logarithmic  and  Exponen- 
tial Functions.     Hyperbolic  Functions        .         .         .  79-87 

CHAPTER  VIII 

Formulas  for   Differentiation  of  Trigonometric   and  Anti- 
trigonometric  Functions         .         .        .        .         .        .  88-95 

CHAPTER  IX 

Rolle's  Theorem.     Law  of  the  Mean.     Taylor's  Theorem    96-111 

vii 


Vlll  CONTENTS 

CHAPTER   X 

PAGES 

Binomial  Theorem 112-115 

CHAPTER   XI 

Differentials  .         . .         .      116-118 

CHAPTER   XII 

Slope  of  the  Tangent  Line  in  Polar  Coordinates.      Subtan- 

gent.     Subnormal.     Asymptotes 119-128 

CHAPTER   XIII 

Curvature.     Eadius  of  Curvature.     Evolutes  and  Involutes 

129-144 

CHAPTER   XIV 

Differentiation  of  Fdnctions  of  Two  Independent  Variables 

145-152 

CHAPTER   XV 
Integration 153-158 

CHAPTER  XVI 
Plane  Areas  by  Indefinite  Integrals         ......      159-168 

CHAPTER   XVII 
Methods  of  Integration 169-180 

CHAPTER  XVIII 
Reduction  Formulas 181-187 

CHAPTER   XIX 

Summation  of  f(x)  -Ax.      Plane  Areas  in  Rectangular  Co- 
ordinates .  ........      188-204 

CHAPTER  XX 

Summation  of  \  {/(0)}2A0.     Plane   Areas  in  Polar  Coordi- 
nates .  205-209 


CONTENTS  IX 

CHAPTER   XXI 

PAGES 

Theorem  in  Infinitesimals.     Definite  Integral  in  General    210-214 

CHAPTER   XXII 


Summation   of    VAx2  +  Ay'2.      Length   of   an  Arc    of  a  Plane 

Curve 215-224 

CHAPTER   XXIII 
Areas  of  Surfaces  of  Revolution       .         .         .         .<  225-229 

CHAPTER   XXIV 
Volumes  by  Means  of  Parallel  Cross  Sections        .         .      230-236 

CHAPTER   XXV 
Successive  Integration 237,  238 

CHAPTER  XXVI 
Plane  Areas  by  Double  Integration  .....      239-244 

CHAPTER   XX VII 
Volumes  of  Revolution  by  Double  Integration       .         .      245-251 

CHAPTER   XXVIII 
Volumes.     Surfaces         . 252-262 

CHAPTER   XXIX 
Some  Methods  of  Approximate  Integration       .         .         .      263-277 

CHAPTER   XXX 
Elements  of  Kinematics 278-287 

CHAPTER   XXXI 
Force.     Mass.     Equation  of  Motion  of  a  Particle  .         .      288-295 

CHAPTER   XXXII 
Rectilinear  Motion 296-308 


X  CONTENTS 

CHAPTER   XXXIII 

PAGES 

Motion  in  a  Plane  Curve 309-316 

CHAPTER   XXXIV 
Work  and  Energy .         .      317-328 

CHAPTER  XXXV 

Attraction       .         .         .         .        .  -       .        .        .         .         .      329-338 

CHAPTER   XXXVI 
Centers  of  Gravity 339-355 

CHAPTER   XXXVII 
Moments  of  Inertia 356-362 

Index 363-364 


THE  ELEMENTS   OP  THE 
DIFFERENTIAL  AND  INTEGRAL  CALCULUS 


DIFFERENTIAL  AND  INTEGRAL 
CALCULUS 

INTRODUCTION 

1.  Constants  and  variables.  A  number  that  retains  the  same 
value  throughout  any  given  problem  is  called  a  constant. 

A  number  that  changes  from  one  value  to  another  in  the 
same  problem  is  called  a  variable. 

For  example,  in  the  equation  of  the  circle  x2  +  y2  =  a2,  a  is  a 
constant  and  x  and  y  are  variables. 

2.  Functions.  A  number  that  so  depends  on  a  second  num- 
ber for  its  value  that  when  the  second  is  made  to  assume 
different  values  it,  in  general,  assumes  different  values,  is  said 
to  be  a  function  of  the  second  number. 

For  example,  a2  —  x2  is  a  function  of  x  because,  as  x  is  made 
to  assume  different  values,  a2  —  x2,  in  general,  assumes  different 
values. 

It  is  said  in  general  because  sometimes  for  different  values 
of  the  variables  the  function  has  the  same  value. 

For  example,  in  a2  —  x2,  for  a  value  of  x  with  the  minus  sign 
affixed  and  the  same  value  of  x  with  the  plus  sign  affixed  the 
function  has  the  same  value. 

3.  From  the  definition  of  a  function  it  is  seen  that  any  com- 
bination of  letters  used  to  denote  a  number  is  a  function  of 
these  letters.  It  may  be  called  a  function  when  the  thought  is 
of  the  combination  of  letters,  and  a  number  when  the  thought 
is  of  the  value  which  this  combination  represents. 

B  1 


2  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

4.  By  a  combination  of  letters  is  meant  an  expression  which 
actually  contains  these  letters  and  cannot  be  reduced  to  an 
equivalent  expression  which  does  not  contain  them. 

Thus,  x*  —  y2  is  a  combination  of  x  and  y,  since  it  contains 
x  and  y  and  cannot  be  reduced  to  an  equivalent  expression  that 

does  not  contain  them.     The  expression  (2+y)(2— y)'+y\y-\ — 

is  not  a  combination  of  x  and  y  because  on  simplification  it 
reduces  to  4  -f  x,  which  does  not  contain  y. 

5.  A  combination  of  letters  is  not  a  function  of  a  letter  not 
actually  contained  in  the  combination.  This  is  obvious,  be- 
cause the  expression  evidently  does  not  assume  different  values 
as  the  letter  assumes  different  values. 

Thus,  (2  +  y)  (2  —  y)  +  y(  y  +  -  J  is  not  a  function  of  y. 

6.  Independent  and  dependent  variables.  In  an  equation  con- 
taining two  variables,  if  one  of  the  variables  is  given  a  value 
chosen  at  pleasure,  the  other  cannot  be  given  a  value  chosen  at 
pleasure.  It  assumes  the  value  got  by  solving  the  equation. 
The  one  to  which  a  value  was  given  at  pleasure  is  called  the 
independent  variable.  The  one  that  assumes  the  value  got  by 
solving  the  equation  is  called  the  dependent  variable. 

For  example,  in  the  equation  x2  +  y2  =  a2,  suppose  that  x  is 

given  the  value  -  •     Then  y  must  be  such  that  —  -f-  y2  =  a2,  or 

n         , — 

y  =  ±  -  V 3.     Here  x  is  the  independent  variable  and  y  the 

dependent  variable. 

It  will  be  noticed  that  independent  variable  is  merely  an- 
other name  for  variable,  and  dependent  variable  another  name 
for  function. 

7.  Classification  of  functions.  It  is  convenient  for  our  pur- 
pose to  make  a  rough  classification  of  functions  as  follows : 

Algebraic  functions.  Those  in  which  the  only  operations 
performed  upon  the  variable  are  a  definite  number  of  opera- 


INTRODUCTION  3 

tions  of  addition,  subtraction,  multiplication,  division,  extrac- 
tion of  roots,  or  raising  to  powers. 

Logarithmic  functions.  Those  involving  the  logarithm  of  the 
variable  or  of  an  algebraic  function  of  the  variable. 

Exponential  or  anti-logarithmic  functions.  Those  involving 
an  exponential  function  of  the  variable  or  of  an  algebraic 
function  of  the  variable. 

Trigonometric  functions.  Those  involving  a  trigonometric 
function  of  the  variable  or  of  an  algebraic  function  of  the 
variable. 

Anti-trigonometric  functions.  Those  involving  an  anti-trigo- 
nometric function  of  the  variable  or  of  an  algebraic  function 
of  the  variable. 

We  may  have  combinations  of  these  functions,  as,  for  ex- 
ample, log10  sin  (tan-1  x2) ;  but  there  would  be  no  advantage  in 
classifying  such  functions,  and  we  shall  not  do  so. 

8.  Algebraic  functions.  An  algebraic  function  in  any  num- 
ber of  unknowns  may  be  integral  or  fractional,  rational  or 
irrational. 

An  integral  algebraic  function  is  one  in  which,  after  all  the 
negative  exponents  have  been  made  positive,  there  are  no 
variables  in  the  denominator. 

Example.  The  expression  V#  +  2  xy  -f-  3  y2  +  Vz  is  an 
integral  algebraic  function. 

A  fractional  algebraic  function  is  one  in  which,  after  all  the 
negative  exponents  have  been  made  positive,  there  are  vari- 
ables in  the  denominator. 

Example.  The  expression  s/x  -f-  2  xy  +  —  -\-  -\/z  is  a  frac- 
tional  algebraic  function. 

A  rational  algebraic  function  is  one  in  which  the  variables 
are  all  raised  to  powers  whose  exponents  are  integers,  positive 
or  negative. 


4  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Examples.  The  expressions  x2  +  y  and  x2  +  y~2  —  4  are 
rational  algebraic  functions. 

An  irrational  algebraic  function  is  one  in  which  at  least  one 
power  whose  exponent  is  a  fraction  appears  somewhere  over 
the  unknowns. 

Examples.  The  expressions  x2-\-y^  and  ^/x2  -\-y2  ±xz  are 
irrational  algebraic  functions. 

9.  It  must  be  noticed  that  the  character  of  the  function  is 
determined  by  the  manner  in  which  the  variables,  not  the 
constants,  are  involved  in  the  function. 

Thus,  ax2  H xy*  +-y2  is  integral  if  b  and  c  are  constant ; 

■Vb  c 

fractional  if  b  or  c  is  variable  ;  rational  if  b  and  y  are  constant; 
irrational  if  b  or  y  is  variable. 

10.  Variables  involved  explicitly  ;  implicitly.  In  an  equation' 
in  two  variables,  either  variable  is  said  to  be  involved  explicitly 
when,  for  a  given  value  of  the  other,  its  value  can  be  found 
without  solving  an  equation.  It  is  said  to  be  involved  im- 
plicitly when,  for  a  given  value  of  the  other,  its  value  can  be 
found  only  by  solving  an  equation. 

For  example,  in  the  equation  y  =  x2  +  4  x  +  1,  let  x  be  given 
some  value.  Then  the  value  of  y  corresponding  to  that  value 
of  x  can  be  found  without  solving  an  equation.  Then  y  is 
involved  explicitly  in  the  equation.  If  y  is  given  some  value, 
the  corresponding  value  of  x  can  be  found  only  by  solving  the 
equation  x2  -f  4  x  ==  y  —  1.  Then  x  is  involved  implicitly  in  the 
equation. 

11.  It  is  generally,  though  not  always,  of  advantage  to  arrange 
the  variables  so  that  the  dependent  variable  is  involved  explic- 
itly in  the  equation.  Thus,  in  the  equation  y  =  sin  x,  if  y  is 
regarded  as  the  dependent  variable,  the  equation  would.be 
written  in  the  above  form,  while  if  x  is  regarded  as  the  depend- 


INTRODUCTION  5 

ent  variable,  the  equation  would  most  naturally  be  written  as 
x  =  sin-1  ?/.  Or  again,  in  y  =  x2,  if  y  is  regarded  as  the  dependent 
variable,  the  equation  would  be  written  in  the  above  form,  while 
if  x  is  regarded  as  the  dependent  variable,  the  equation  would 
most  naturally  be  written  asx=±  Vy. 

12.  Functional  symbols.  A  function  may  be  denoted  by  any 
convenient  symbol,  as  /,  F,  <j>,  U,  •  ••,  followed  by  the  variable 
or  variables  inclosed  in  brackets.  If  there  are  more  variables 
than  one  in  the  function,  the  variables  in  the  symbol,  in  general, 
are  separated  by  commas.  It  is  said  in  general  because  in  a 
few  rare  cases  they  enter  in  a  certain  combination  only.  In 
these  cases  they  may  or  may  not  be  separated  by  commas. 

For  example,  a?2 +4  may  be  denoted  by  f(x)  ;  x2+3xy-\-2y2+6 
by  f(x,  y),  and  (xy)2  +  4  (xy)  -f- 1,  in  which  x  and  y  appear  in 
the  combination  xy,  by  f(xy)  or  f(x,  y). 

During  the  investigation  of  any  particular  problem  the  same 
functional  symbol  must  be  used  to  denote  the  same  operation 
or  series  of  operations. 

For  example,  if  f(x)  =  x2  +  4,  then  f(y)  =  y2  -f  4,  and 
f(xy)  =  (xy)2  +  4. 

13.  Absolute  value  of  real  function.  By  the  absolute  value  of 
a  real  function  is  meant  the  value  of  the  function  disregarding 
the  minus  sign  if  the  function  is  negative. 

1  1 

For  example,  the  absolute  value  of  -  when  x  =  —  2  is  — 

x  2 

Absolute  value  is  denoted  by  two  parallel  lines.      Thus, 


14.   Single-valued,  double-valued,  multiple-valued  functions.     A 

mnction  of  one  variable  is  said  to  be  single- valued  when  for 
a  value  of  the  variable  the  function  has  just  one  value. 

Thus,  in  the  function  given  by  the  equation  y  =  log10  x,  for  a 
value  of  x,  y  or  \og10x  has  just  one  value.  Then  y  or  log10a;  is 
a  single-valued  function  of  x. 


6 


DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


In  the  curve  whose  equation  in  Cartesian  coordinates  gives 
a  single-valued  function  of  x,  any  line  parallel  to  the  y-axis  will 

cut  the  curve  in  just  one 
point  or  not  at  all. 

Thus,  in  the  curve  y  = 
log10aj  (see  Fig.  1),  any  line, 
as  AB  or  CD,  parallel  to  the 
2/-axis,  cuts  the  curve  in  one 
point  or  not  at  all. 

A  function  of  one  varia- 
ble is  said  to  be  double- 
valued  when,  for  a  value  of 


Fig.  .  1. 

the  variable,  the  function  has  two  values. 


Thus,  in  the  equation  x2  +  y2  =  4,  or  y=± V4  —  x2,  for  a 
value  of  x,  y  or  ±  V4  —  x2  has  two  values.  Then  y  or 
±  V4  —  x2  is  a  double-valued  function  of  x. 

In  the  curve  whose  equation  in  Cartesian  coordinates  gives 
a  double-valued  function  of  x,  any  line  parallel  to  the  y-axis 
will  cut  the  curve  in  two  points,  distinct  or  coincident,  or  not 
at  all. 

Thus,  in  the  curve  y  = 
±  V4  —  x2  (see  Fig.  2),  any 
line,  as  AB  or  CD,  parallel 
to  the  ?/-axis,  cuts  the  curve 
in  two  points,  distinct  or 
coincident,  or  not  at  all. 

A  function  of  one  varia- 
ble  which    has    more    than 
two   values    for   a   value   of 
the  variable  is  called   a  multiple-valued  function  of  the  vari- 
able. 

Thus,  in  the  equation  y  =  tan"1  x,  y  or  tan"1  x  has  more  than 
two  values  for  a  value  of  x.  Then  y  or  tan"1  a?  is  a  multiple- 
valued  function  of  x. 

In  the  curve  whose  equation  in  Cartesian  coordinates  gives 


Fig.  2. 


INTRODUCTION 


B 

Y 

D 

37T   ^ 

TT            

0 

X 

-7T 

Ia 

C 

a  multiple-valued  function  of  x,  any  line  parallel  to  the  ?/-axis 
will  usually  cut  the  curve  in  more  than  two  points. 

Thus,  in  the  curve  y  = 
tan-1  re  (see  Fig.  3),  any  line, 
as  AB  or  CD,  parallel  to  the 
2/-axis,  cuts  the  curve  in  an 
unlimited  number  of  points. 

15.  In  any  particular  prob- 
lem it  is  desirable  to  have  the 
function  f(x)  single-valued. 
If  it  is  not  single-valued,  it 
can  be  made  so  by  considering  it  as  made  up  of  a  number  of 
functions,  each  of  which  is  single-valued.  The  curve  whose 
equation  gives  the  function  will  then  consist  of  a  number  of 
branches,  each  of  which  will  correspond  to  one  of  the  com- 
ponent functions  of  fix). 

Thus,  in  the  equation  y=±  V4  —  x2,  the  function  ±  V4  —  x2 
may  be  considered  as  made  up  of  the  two  functions  +  V4  —  x2 
and  —  V4  —  x2,  each  of  which  is  single-valued.  In  the  curve 
y  =  ±  V4  —  x2  (see  Fig.  2),  the  part  above  the  ic-axis  corre- 
sponds to  y  =  +  V4  —  x2,  and  the  part  below  to  y  =  —  V4  —  x2. 


Fig.  3. 


EXERCISES 

1.  For  what  values  of  n  does  xn(n~1}  cease  to  be  a  function 
of  x? 

2.  In    the    equation   y(l+    +fl? )  =  (  z  +  x-\ — ^-  ]    show 

\        z  —  xj      \  z  —  xj 

that  y  is  a  function  of  z  but  not  of  x. 

3.  Show  that  sin  x  tan  i  x  +  cos  x  is  not  a  function  of  x. 

4.  In  the  equation  xy  —  2  x  +  y  =  n,  determine  the  value 
of  n  for  which  y  ceases  to  be  a  function  of  x. 


8  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

In  each  of  the  exercises  5  to  9  inclusive,  express  y  explicitly 
in  terms  of  x  and  a. 


a2 
5.    log10x  +  log10y  —  21og10a  =  0.  Ans.  y  =  — 


x 

6.  3  log10  x  —  2  log10  y  -f  4  log10  a  =  0.       J.ns.  y  =  ±  Va;3a4. 

7.  sin-1  a>  +  sin-1  y  =  a.     Ans.  y  —  Vl  —  x2  sin  a  —  x  cos  a. 

8.  tan"1  a  +  tan-1  ?/  =  a.  ^s.  y=  tana~a;  . 

1  +  x  tan  a 

9.  cos_1ic  -f-  tan-1  y  =  a.  Ans.  y  = 


as  +  Vl  —  x2  tan  a 

10.  If f(x)  ee  ar5-  3  a^+  6  a;  +  5,  prove  that /(l)  =  9,/(3) =23, 
/(-I)  =-5. 

11.  If  /(a;,  y)  =  x2  +  2  a?y  +  y2,  find  /(y,  a?)  and  f(x,  x). 


12.  If  /(a?)  =  Vl  —  a?2,  find  /(sin  a?)  and  /(cos  a?). 

13.  If    /(£)ee£2  +  4     and     ^(aj)  =  (x  - 1),    find    f(F(x)), 
F(f(x)),  f(f(x)),  F(F(x)). 

14.  If  f(x,  y)  =  ^1,  find  f(x,  y)  +f(y,  x). 

x  -\-y 

15.  If  f{x)  =  ^,  prove  that  /^"^  =  £=£. 

a?  +  1  1  +f(x)f(y)      1  +  xy 


16.  If  <£(aj)=Vl-a?,  find  ^(Vl-ar2). 

17.  If  /(as)  =  tana;,  find  /^.V^l         ^s-  tan(a?'  +  v). 

W  1-/0*0/(30  ^         ; 

18.  If  /(a?)  =  sin  a?  and  $  (a?)  =  cos  x,  find 

<£  (»)  <k  (y)  T  f(x)f{y).  Ans.  cos  (a?  ±  ?/). 

19.  If /(a?)  =  since  and  <£(a?)  ==  cos  a?,  find 

/(«)  4>(2/)±4>  (»)/(y).  -4w«.  sin  (a>  ±  ?/). 

20.  If  F(a;)  =  log10a3,  prove  that  F(a?)  +  F(y)  =  F(xy)  and 
F(an)  =  nF(x). 


INTRODUCTION 

21.  If  cf)  (x)  =  ax,  prove  that  <f>  (a?)  <£  (y)  =  <f>  (a?  +  y), 
<£  (a;)  -t-  cfi  (y)  =  <fi(x  —  y)  and  <£  (na;)  =  \  $  (x)  \ ". 

22.  If  JF^a;)  =  cos  2  a:,  prove  that 

F(a>)  +  F(y)  =  2  *p±*) F(^y 

23.  If  /(a?,  ?/)  =  Ax  +  £#  +  C,  show  that  /(a;,  y)  =  0  and 
f(—y,x)  =  0  are  the  equations  of  two  straight  lines  perpen- 
dicular to  each  other. 

24.  If  f(x,  y)  =  2x  +  3y-±,  what  do  /(a2, 2/)  =  0,  f(x,  y*)  =  0, 
f(x2,  y2)  =  0,  f(x2,  0)  =  0  represent  ?     Plot  the  curves. 

25.  Find  |  x2  —  x  |  when  #  =  i. 

I  sin  a?  I     when  a;  =  210°. 
I  logio #  J  when  aj  =  .001. 

26.  Find  |a?3  —  a^+'a;|  when  x  =  —  a. 

\x* -\-x2  —  x\  when  x  =  —  a. 


27.  Plot  the  curve  y  =  ±  Vl  —  (a?  —  l)2.  Indicate  which 
part  of  the  curve  is  found  by  taking  y  =  +  Vl  —  (x  —  l)2, 
and  which  by  taking  y  =  —  Vl  —  (a;  —  l)2. 


28.  Plot  the  curve  y  =  1  ±  V2  —  (a?  —  3)2.  Indicate  which 
part  of  the  curve  is  found  by  taking  y=l  +  V2  —  (x— 3)2,  and 
which  by  taking  y=l  —  V2  —  (x— 3)2. 


CHAPTER   I 

LIMITS 

16.  Definitions.  To  determine  whether  one  number  is  greater 
or  less  than  another,  represent  the  two  numbers  as  lengths  from 
the  origin  on  a  straight  line.  The  one  represented  by  the 
length  whose  extremity  lies  farther  to  the  right  is  the  greater 
of  the  two. 

For  example,  to  determine  whether  —  2  is  greater  or  less 
than  — 10,  represent  the  two  numbers  as  lengths  from  the  ori- 
gin on  a  straight  line  (Fig.  4). 

—10  .    -2        o       The   extremity   of   the   line 

Fig.  4.  1  «  -, . 

that  represents  —  2  lies  far- 
ther to  the  right.     Then  —  2  is  greater  than  —  10. 

A  variable  or  function  is  said  to  continually  increase  in  value 
when  it  assumes  a  succession  of  values  each  of  which  is  greater 
than  the  preceding  value. 

A  variable  or  function  is  said  to  continually  decrease  in  value 
when  it  assumes  a  succession  of  values  each  of  which  is  less 
than  the  preceding  value. 

17.  Let  Sn  denote  the  sum  of  the  first  n  terms  of  the  series : 

1+1+1+1+.... 

^2     22     23 
Then 


Sn 

=1+-+—+ 
^  2      22 

1 

23 

+ 

1- 

(IT 

=  1- 

1 

\   / 
1 

2 

=  2 

-2| 

$r- 

:2 

2»-i 


n-l 


LIMITS  11 

Since  n  denotes  the  number  of  terms  taken,  it  must  be  a 
positive  integer,  and  since  any  number  of  terms  may  be  taken, 
it  may  be  any  positive  integer.  Start  with  a  value  of  n,  say 
n  =  1.  Let  n  continually  increase,  and  see  what  change  is 
being  produced  in  Sn. 

By  taking  n  large  enough,  ■— -  can  be  made  less  than  any 

positive  number,  8,  we  may  choose  to  name,  no  matter  how 

small  a  number  is  chosen  for  8. 

1       1 

For  example,  if  8  is  chosen  to  be  — ,  can  be  made  less 

r    >  79'  2n~1   ' 

than  —  by  taking  n  equal  to  8. 

Also,  — —  remains  less  than  8  for  all  values  of  n  subsequent 
to  the  one  chosen  to  make  it  less  than  8.     Then  Sn  =  ? 


On-l' 

where can  be  made  to  come  as  near  the  value  zero  as  we 

2"-1  1 

please  by  taking  n  large  enough,  and  where  — -  remains  at  least 

Zi 

that  near  zero  for  all  subsequent  values  of  n.    Then  with  regard 
to  Sn,  as  n  is  allowed  to  continually  increase,  we  may  state : 

(1)  That  Sn  can  be  made  to  come  as  near  the  value  2  as  we 
please  by  taking  n  large  enough. 

(2)  That  Sn  remains  at  least  that  near  2  for  all  subsequent 
values  of  n. 

18.  That  Sn,  in  the  preceding  article,  comes  and  remains  as 
near  2  as  we  please  can  be  seen  more  clearly  perhaps  by  repre- 
senting its  successive  values  as  lengths  from  the  origin  on  a 
straight  line. 

Let   A'OA   be   a   line    of 

indefinite   length    (see   Fig.    jj       J J ^— jj- 

5).     On  A'OA,  plot  the  line 
OB,  2  units  in  length.     Be- 
ginning with  n  =  1,  plot  the  successive  values  of  Sn  as  lengths 
on  A'OA.       JS1}  or   Sn  when   n  =  l,  =2  —  1  =  1.      Therefore 
the  extremity  of  Sx  is  at  Ax,  which  bisects  OB.     S2,  or  Sn  when 


c|b 


12         DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

n  =  2,  =2  —  1  Therefore  the  extremity  of  S2  is  at  A2,  which 
bisects  AXB.  S3,  or  Sn  when  n  =  3,  =2  —  \.  Therefore  the 
extremity  of  S3  is  at  A3,  which  bisects  A2B.  The  law  which 
the  successive  values  of  Sn  obey  is  now  evident.  It  is :  the 
extremity  of  Sn  is  found  by  bisecting  the  line  An_xB  where 
An_Y  is  the  extremity  of  Sn_Y. 

Since  the  line  with  its  extremity  at  B  is  continually  bisected, 
it  follows  that  if  any  distance  BC  (equal  to  8,  suppose),  no 
matter  how  short,  is  marked  off  from  B  to  the  left,  the  ex- 
tremity of  Sn  will  eventually  fall  between  O  and  B,  and  remain 
between  (7  and  B  for  all  subsequent  values  of  n. 

19.  Definitions.  If  a  variable  is  allowed  to  continually  in- 
crease, becoming  greater  than  any  number  we  may  choose  to 
name,  no  matter  how  great  that  number  may  be,  it  is  said  to 
increase  without  limit  or  to  become  infinite. 

Thus,  n  in  the  problem  of  Art.  18  increases  without  limit  or 
becomes  infinite. 

The  notation  used  to  indicate  that  a  variable  increases  with- 
out limit  or  becomes  infinite  is  n  =  go,  which  is  read,  "  increases 
without  limit "  or  "  becomes  infinite." 

If,  as  the  variable  changes  according  to  a  given  law,  the 
function  comes  as  near  a  fixed  number  as  we  please  and 
remains  at  least  that  near  for  all  subsequent  values  of  the 
variable,  the  function  is  said  to  approach  the  fixed  num- 
ber as  a  limit  as  the  variable  changes  according  to  the  given 
law. 

Thus,  as  n  increases  without  limit,  Sn,  in  the  series  1  +  -  + 

11  i 

— h  — h  •••,  comes  as  near  2  as  we  please,  and  remains  at  least 

02    ■    93 

that  near  2  for  all  subsequent  values  of  n.  Then  jSn  is  said  to 
approach  2  as  its  limit  as  n  increases  without  limit. 

The  notation  used  to  indicate  that  as  n  increases  without 
limit  Sn  approaches  2  as  its  limit  is  ^^  ["$„!  =  2,  which  is 
read,  "  the  limit  of  Sn  as  n  increases  without  limit  is  2." 


LIMITS  13 

11  1 

20.   In  the  example  Sn  =  l  +  -  +  --\ h^^is  always 

less  than  its  limit,  and  approaches  it  by  continually  increas- 
ing to  it.  That  this  is  not  always  the  case  may  be  seen  by 
examples. 

Example  1.     Find  the  limit  which  the  sum  of  the  first  n 
terms  of  the  series  111 

approaches  as  n  increases  without  limit. 

o  =_i_l_L  _i        L 

2      22     23       "      2n~l 


i-i 


=  -2  +  - 

1      f)»-l 


1 

2? 


Therefore  limit  [sJ  =  -  2. 


In  this  example  £w  is  always  greater  than  its  limit  and 
continually  decreases  to  it. 

Example  2.     Find  the  limit  which  the  sum  of  the  first  n 
terms  of  the  series  i      i       1 

2      22     23 

approaches  as  n  increases  without  limit. 


2     22     23  v      J      2n~1 

IV 


-.i-t-' 


2_2/__ly_2      (-l)w+1 
3     3V      2/   ~3      3-2-1" 


14  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Therefore  limit  [Vl  =  ?. 

In  this  example  Sn  is  alternately  greater  and  less  than  its 
limit. 

21.  It  must  be  carefully  noticed  that  it  is  as  essential  that 
the  function  remain  as  near  the  fixed  number  as  we  please  as 
that  it  can  be  made  to  come  as  near  the  fixed  number  as  we 
please. 

Consider  the  series 

2     4     8 
found  by  adding  + 1  and  —  1  alternately  to  the  terms  of  the 

series  i       1       i 

1  +  -  +  -  +  -+— . 

9         92        93 

4J  Zi  Zl 

Let  JSn  denote  the  sum  of  the  first  n  terms  of  the  series. 
Then  #n  =  2-l  +  5_Z+...  +  fJL±l\ 

n  9    '    A         C    '  '        9»-l  r 


2      4      8  V2'1-1 

where  the  plus  sign  must  be  taken  in  the  ?ith  term  if  n  be  odd, 
and  the  minus  sign  if  n  be  even. 

This  series  can  be  summed  as  follows : 

*.=a+i)+g-i)+(i+i)+(i-i)+-+(^±i) 

=1+^+i+|+-+2^+(1-1+1-1-±1)> 

by  removing  brackets  and  rearranging  terms. 

*-| 

.•.  Sn  = r-  +  (0  or  1,  according  as  n  is  even  or  odd). 

l-± 

2 

=  2  —  —,  if  n  is  even, 
or  =  3  —  - — - ,  if  n  is  odd. 

On— 1' 


LIMITS  15 

By  taking  n  large  enough  and  even,  Sn  can  be  made  to  come 
as  near  2  as  we  please,  and  by  taking  n  large  enough  and  odd,  Sn 
can  be  made  to  come  as  near  3  as  we  please.  Then  if  we  start 
with  n  any  positive  integer  and  allow  it  to  increase  by  unity  each 
time,  Sn  can  be  made  to  come  as  near  2  or  3  as  we  please.  It 
does  not,  however,  remain  either  as  near  2  or  3  as  we  please 
for  all  subsequent  values  of  n.  It  therefore  approaches  neither 
2  nor  3  as  a  limit,  and  Sn  as  n  increases  without  limit,  by  unity 
each  time,  approaches  no  limit. 

22.  Although  the  function  considered  in  the  preceding  article 
failed  to  approach  a  limit,  it  always  remained  between  two 
numbers.  Eor  example,  it  remained  between  1  and  3.  Now 
a  function  may  fail  to  approach  a  limit  in  other  ways,  as  the 
following  examples  will  show : 

Example  1.     Consider  the  series 

l+2-f-22  +  23+-... 
£n  =  l  +  2  +  22  +  23+...+  2M-1 

1-2' 


1-2 


2n 


As  n  increases  without  limit,  Sn  becomes  greater  than  any 
number  we  may  choose  to  name.  It  also  remains  greater  for 
all  subsequent  values  of  n.  Therefore  Sn  does  not  come  and 
remain  as  near  any  given  number  as  we  please  and  therefore 
does  not  approach  a  limit. 

Definition.  If,  as  the  variable  changes  according  to  a  given 
law,  the  function  becomes  and  remains  greater  than  any  number 
we  may  choose  to  name,  no  matter  how  great  that  number  may  be, 
it  is  said  to  increase  without  limit  or  to  become  infinite. 

The  notation  used  to  indicate  that  as  n  increases  without 

limit  Sn  increases  without  limit  is  w™1^  \$J\  =oo,  which  is  read, 

"as  n  increases  without  limit,  Sn  increases  without  limit." 
In  Example  1,  Sn  increases  without  limit. 


16         DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Example  2.     Consider  the  series 

_l_2-22-23 . 

Sn  =  -  1  -  2  -  22  -  23 2" 

1        9n 
=  —  1  •  =  1  —  2M. 

1-2 

As  n  increases  without  limit,  Sn  becomes  less  than  any- 
negative  number  we  may  choose  to  name  no  matter  how  great 
in  absolute  value  that  number  may  be.  It  also  remains  less 
for  all  subsequent  values  of  n.  Therefore  Sn  does  not  come 
and  remain  as  near  any  given  number  as  we  please,  and  there- 
fore does  not  approach  a  limit. 

Definition.  If,  as  the  variable  changes  according  to  a  given 
law,  the  function  becomes  and  remains  less  than  any  negative 
number  we  may  choose  to  name,  no  matter  how  great  in  absolute 
value  that  number  may  be,  it  is  said  to  decrease  without  limit 
or  to  become  infinite  negatively. 

The  notation  used  to  indicate  that  as  n  increases  without 

limit  Sn  decreases  without  limit  is    ™^  \Sn  \  =  —  co,  which  is 
read,  "as  n  increases  without  limit,  Sn  decreases  without  limit." 
In  Example  2,  Sn  decreases  without  limit. 

Example  3.     Consider  the  series 

l_2  +  22-23  +  -... 
Sn  =  1  -  2  +  22  -  23  + ...  +  (-  l)*-^""1 

_  1  +  (—  l)n+12n _  (—  1)W2W  —  1 

"         1+2         ~  3 

As  n  increases  without  limit,  Sn  is  such  that  in  absolute 
value  it  becomes  greater  than  any  number  we  may  choose  to 
name.  It  also  remains  greater  in  absolute  value  for  all  subse- 
quent values  of  n.  Therefore  #n  does  not  come  and  remain  as 
near  any  given  number  as  we  please  and  therefore  does  not 
approach  a  limit. 


LIMITS  17 

In  this  case  Sn  is  such  that  in  absolute  value  it  increases 
without  limit. 

The  notation  used  to  indicate  that  as  n  increases  with- 
out  limit,  Sn,  in   absolute  value,  increases   without   limit   is 

limit  r  i$  i"l  _  q-   whica  is  read,  "as  n  increases  without  limit, 
/l  =  CO  |_  '       '  J 

Sn,  in  absolute  value,  increases  without  limit." 

EXERCISES 

1.  In  the  examples  of  Arts.  20,  21,  and.  22,  represent  the 
successive  values  of  Sn  as  lengths  from  the  origin  on  a  straight 
line,  determining  the  law  which  these  values  obey. 

In  each  of  the  following  series  find  the  limit,  if  any,  which 
Sn  approaches  as  n  increases  without  limit.  In  each  case 
represent  the  successive  values  of  Sn  as  lengths  from  the  origin 
on  a  straight  line. 

2-    1+1  +  1  +  1+....  4.    1-1  +  1-1  +  -. 

6.    J_  +  J_  +  J_  +  J_  +  .... 
1  •  2     2  •  3     3  •  4     4  •  5 


Suggestion. 


1-2     2-3      3-4     4-5  rc(w-l)     V       2/      \2      3 


3     4J 


+      T-^      +-+-- 


U     b)  \n     n-1 

2T4T8     16  2      4^8 

23.  Definitions.  If  a  variable  is  allowed  to  continually  de- 
crease becoming  less  than  any  negative  number  we  may  choose  to 
name,  no  matter  how  great  in  absolute  value  that  number  may 
be,  it  is  said  to  decrease  without  limit  or  to  become  infinite 
negatively. 


18  DIFFERENTIAL  AND  INTEGRAL    CALCULUS 

The  notation  used  to  indicate  that  a  variable  decreases  with- 
out limit  or  becomes  infinite  negatively  is  —  oo ,  which  is  read, 
"  decreases  without  limit  or  beco2nes  infinite  negatively." 

If  a  variable  assumes  a  succession  of  values  each  of  which  is 
nearer  a  given  number  than  the  preceding  value,  and  is  made  to 
differ  from  the  given  number  by  a  value  as  small  as  we  please,  it  is 
said  to  approach  the  given  number  as  a  limit. 

The  notation  used  to  indicate  that  a  variable  approaches  a 
given  number  a  is  =  a  which  is  read,  "approaches  a  as  a 
limit." 

Note.  Attention  should  be  called  to  the  notation  =  and  =.  When 
the  variable  approaches  a  given  number,  we  use  the  notation  =,  and  when 
it  increases  or  decreases  without  limit,  we  use  the  notation  =. 

24.  In  all  problems  in  limits  we  shall  suppose  that  the  vari- 
able changes  according  to  one  or  other  of  the  following  laws : 

1st.  It  increases  without  limit. 
2d.  It  decreases  without  limit. 
3d.   It  approaches  a  given  number  as  a  limit. 

When  the  variable  changes  according  to  one  of  these  laws,  it 
may  assume  an  unlimited  number  of  successions  of  values,  the 
only  restriction  being  that  the  values  it  assumes  are  in  accord- 
ance with  this  law. 

Thus,  if  the  variable  increases  without  limit,  it  may  assume 
an  unlimited  number  of  successions  of  values,  the  only  restric- 
tion being  that,  in  any  succession,  each  value  is  greater  than 
the  preceding  and  that  eventually  the  values  become  greater 
than  any  number  we  may  choose  to  name. 

For  example,  it  may  assume  any  one  of  the  successions : 


or 


1, 

91 

6, 

v  2, 

13, 

iej,      • 

1 

If 

3 

1, 

li> 

H, 

If,        • 

1, 

2, 

3, 

4, 

5, 

6,        • 

1, 

3, 

5, 

7, 

9, 

11,        • 

1, 

2, 

4, 

8, 

16, 

32,        • 

LIMITS  19 

If  in  addition  to  the  given  law  the  variable  is  further 
restricted,  some  of  these  successions  become  no  longer  possible. 
Thus,  for  example,  if  the  variable  denotes  the  number  of  terms 
taken  in  a  series,  the  first  two  of  the  above  successions  become 
no  longer  possible. 

If  the  function  approaches  a  limit  when  the  variable  changes 
according  to  any  one  of  the  above  laws,  and  is  not  otherwise 
restricted,  it  will  approach  the  same  limit  when  the  variable 
is  further  restricted.  This  is  obvious,  because  if  the  function 
approaches  a  limit  for  any  succession  of  values  subject  to  a 
given  law,  it  will  also  approach  the  same  limit  when  the 
variable  is  limited  to  fewer  of  these  successions. 

In  the  theorems  of  Arts.  26,  27,  and  28,  the  variable  is  sup- 
posed to  change  according  to  one  or  other  of  the  above  laws, 
and  is  not  otherwise  restricted.  By  the  preceding  paragraph, 
the  theorems  will  also  be  true  if  the  variable  be  further 
restricted. 

As  an  example  in  which  the  function  approaches  a  limit  as 
the  variable  approaches  a  given  number,  consider  the  function 
x  + 1  as  x  approaches  2. 

By  giving  x  a  value  near  enough  to  2,  x  + 1  can  be  made  to 
come  as  near  the  value  3  as  we  please.  Also,  x  + 1  remains  at 
least  that  near  3  for  all  subsequent  values  of  x.     Therefore 

To  illustrate  by  points  on  a  line,  choose  some  succession  of 
values,  suppose  1,  2J-,  If,  2\,  •••. 

As  x  takes  each  of  these  values  in  succession,  x  + 1  takes 

each  of  the  values  2,  3J,  2f,  3|,  •••  in  succession  (see  Fig.  6). 

If  we  mark  off  a  length  of 

5 
line    GB    (equal    to    8),    no  /p 

|C  B 

matter  how  small,  about  the   -+ j * f]  V\  J  . 

point  which   is   three   units  F  2%  3% 

from  the   origin,  x  +  1  will 

eventually  take  a  value  such  that  the  extremity  of  the  line 
that  represents  it  comes  and  remains  in  the  region  GB. 


20  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

25.    Definition.      An  infinitesimal  is  a  function  that  can  be 
made  to  approach  zero  as  a  limit. 

Example   1.      In   the   example   of   Art.    17,   Sn=2 


Since  limit 
n=ao 


r  i 


—  0?  —  9^i   is   infinitesimal   as   n   becomes 


2»-i 

infinite,  being  integral. 

Example  2.  Since  ™?q  x  — 1=0,  #  —  1  is  infinitesimal 
as  x  approaches  1. 

26.  Erom  the  definition  of  a  limit,  the  difference  between  a 
function  and  its  limit  is  infinitesimal. 

Example  1.     In  Sn  =  2 -, is  the  difference  be- 

tween  Bn  and  its  limit  2,  and  is  therefore  infinitesimal  as  n 
increases  without  limit. 

Example  2.     Since  h^  \x  +  ll  =  3,  the  difference  between 

x  + 1  and  its  limit  3  is  infinitesimal  as  x  approaches  2. 

27.  If  the  difference  between  a  funcfciai]  qr>^  a  constarit  ^ 
infinitesimal,  the  constant  is  the  limit  of  the  function. 

Eor,  let  U  be  a  function  and  a  a  constant  such  that  JJ—a=e 
where  e  is  infinitesimal.  Then  U=  a  +  e.  Since  e  is  infini- 
tesimal, U  can  be  made  to  come  and  remain  as  near  a  as  we 
please.     Therefore  £7  approaches  a  as  its  limit. 

This  theorem  is  the  converse  of  that  of  the  preceding  article. 

28.  The  following  theorems  are  necessary  to  the  subsequent 
investigation : 

Theorem  I.  Any  finite  multiple  of  an  infinitesimal  is  an 
infinitesimal. 

Let  e  be  an  infinitesimal  and  m  any  number.  To  prove  that 
me  is  infinitesimal. 

Proof.       Choose  any  number  e'  as  small  as  we  please.    Allow 


e  to  become  less  than  —  •     Then  me  is  less  than  e'.     Since  e  is 

m 


LIMITS  21 

e' 

infinitesimal,  it  remains  less  than  —  for  all  subsequent  values 

of  the  variable.  Therefore  me  can  be  made  to  become  and 
remain  less  than  e\  Therefore  me  can  be  made  to  become  and 
remain  as  small  as  we  please.     It  is  therefore  infinitesimal. 

Example.     If  e  = is  infinitesimal,  lOOOf- — -  ]  is  infini- 

tesimal.  v 

A  special  case  of  this  theorem  is  the  following : 
A  fraction   whose   numerator   is   infinitesimal,   and   whose 
denominator  is  any  number  not  zero,  is  infinitesimal. 

Example.  Since  lva^  \x  —  ll  ==  0.  .-.  ^— =  is  infinitesimal 
as  x  approaches  1. 

Theorem  II.  The  sum  of  a  finite  number  of  infinitesimals  is 
infinitesimal. 

Let  e1?  e2,  e3,  •••,  en  be  n  infinitesimals.  To  prove  that 
3  =  ex  -f-  e2  -f-  e3  +  •  •  •  +  en  is  infinitesimal. 

Proof.  Since  each  of  the  infinitesimals  may  be  made  to 
become  and  remain  as  small  as  we  please,  let  us  make  each 
to  become  and  remain  smaller  in  absolute  value  than  some 
positive  infinitesimal  e.  Then  S  becomes  and  remains  less 
than  ne.  But  ne  is  infinitesimal,  by  Theorem  I.  Therefore 
5  is  infinitesimal. 

Note.  Attention  must  be  called  to  the  necessity  for  the  word  finite  in 
the  above  theorems.  Suppose  that  a  line  of  given  length  is  divided  into  n 
equal  parts  by  bisecting  the  line  and  continually  bisecting  each  part.  If 
n  is  allowed  to  increase  without  limit,  each  part  is  infinitesimal.  The 
sum  of  the  infinitesimals,  however,  is  not  infinitesimal.  It  is  the  length 
of  the  given  line.  In  this  case  there  is  not  a  finite  number  of  infinitesi- 
mals, but  a  number  that  increases  without  limit  as  each  part  approaches 
zero  as  its  limit. 

Theorem  III.  If  two  functions  differ  only  by  an  infinitesi- 
mal, and  one  of  them  approaches  a  limit,  the  other  approaches 
the  same  limit. 


22  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Let  Ux  and  U2  be  two  functions  differing  only  by  an  infini- 
tesimal, and  suppose  that  Ux  approaches  a  as  its  limit.  To 
prove  that  U2  also  approaches  a  as  its  limit. 

Proof.        By  supposition,  C7i—  U2=e,  where  e  is  infinitesimal. 

By  the  definition  of  a  limit,  JJ1  —  a  =  e',  where  e'  is  infini- 
tesimal. 

Subtract.     .*.  U2  —  a  =  e'  —  e. 

Since  e'  and  e  are  infinitesimals,  e'  —  e,  by  Theorem  II,  is 
infinitesimal. 

Therefore,  by  the  theorem  of  Art.  27,  U2  approaches  a  as  its 
limit. 

Theorem  IV.  If  each  of  a  finite  number  of  functions  ap- 
proaches a  limit,  the  limit  of  the  sum  of  the  functions  is  the 
sum  of  their  respective  limits. 

Let  Ulf  U2,  U3,  "•,  Un  be  n  functions  which  approach  the 
limits  aj,  a2,  a3,  •••,  an,  respectively.  To  prove  that  the  limit 
of  £7i  +  U2  +  U3  H h  Un  is  a^atz  +  a^ \-an. 

Proof.        By  supposition, 

U1  =  a1  +  e» 

(J 2  =  (X2  -\-  €2, 

c/3  =  a3  -J-  e3, 


un  —  ctn  +  cn, 

where  ex,  e2,  e3,  •••,  en  are  infinitesimals. 

.-.  Ui+Uz-t-  U8-\ h  Un=(al  +  a2  +  a3-\ \-an) 

■f  (ei-f  e2  +  e3+  •••€„). 

Now  ex  -f-  e2  +  e3  +  •  ••  +  en  is  infinitesimal,  by  Theorem  II. 

Therefore    £7i  -f-  J72  -f  Us  + h  Un   approaches    %  +  a2  +  a 

+  "•  +an  as  a  limit,  by  the  theorem  of  Art.  27. 

Example.     ^[^-^'"M+Sf-3]. 


LIMITS  ■  23 

Theorem  V.  If  each  of  a  finite  number  of  functions  ap- 
proaches a  limit,  the  limit  of  the  product  of  these  functions 
is  the  product  of  their  respective  limits. 

Let   Ij\,  U2,  U&  ••-,  Un  be  n  functions  which  approach  the 
limits  aly  a2,  as,  •  ••,  an  respectively.     To  prove  that  the  limit  of 
Ui'U2'U3"*  Un  is  ftx  •  a2  •  a3  •  ••  an. 
Proof.        By  supposition, 

h\  =  aY  +  fi, 
U2=a2  +  e2, 

U3  =  a3  H-  €3) 


Un  =  Cln  -f-  €nf 

where  e1?  e2,  e3,  •••,  en  are  infinitesimals. 

.'.  Ui'  U2  -  Uz  •••  Un  =  aL  •  a2  •  a3  •••  an  plus  a  finite  number  of 
terms,  each  containing  an  infinitesimal  factor.  Each  of  these 
factors  is  infinitesimal,  and  therefore,  by  Theorem  II,  their 
sum  is  infinitesimal.  Therefore,  by  the  theorem  of  Art.  27, 
Ui  •  U2  -  U3  •••  Un  approaches  ax  •  a2  •  as  •••  an  as  a  limit. 

Example.     »f  *  [*  .  rf]  =  ^  [*>]  .  ^  [<*]. 

Theorem  VI.  If  the  numerator  of  a  fraction  approaches  the 
limit  zero  and  the  denominator  approaches  a  limit  other  than 
zero,  the  fraction  approaches  zero  as  its  limit. 

Let  —  be  a  fraction  in  which  the  limit  of  U  is  0  and  the 

V  U 

limit  of  V  is  b,  where  b  is  not  zero.     To  prove  that  ^  ap- 
proaches zero  as  its  limit. 

Proof.  Since  V  approaches  the  limit  b,  \V\  can  be  made  to 
become  and  remain  larger  than  some  positive  number  c.     Then 

U 


for  all  subsequent  values  of  the  variable, 


V 


is  less  than  —  • 


Approaches 


But ' —  approaches  zero  as  its  limit.     Therefore 

C  ■  U  V 

zero  as  its  limit.     Therefore  —  approaches  zero  as  its  limit. 


Example.     limi| 
x=l 


"x  —  1" 
x  +  2 


=  0. 


24         DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Theorem  VII.  If  the  numerator  of  a  fraction  approaches  a 
limit,  and  the  denominator  approaches  a  limit  other  than  zero, 
the  limit  of  the  fraction  is  the  limit  of  the  numerator  divided 
by  the  limit  of  the  denominator. 

Let  —  be  a  fraction  in  which  the  limit  of  U  is  a,  and  the 

TT 
limit  of  V  is  b,  where  b  is  not  zero.    To  prove  that  —  approaches 

.    a  ' 

the  limit  T» 
b 

Proof.       TJ—  a  +  e, 

V=  b+  fi,  where  e  and  /?  are  infinitesimals. 

"""   V~b  +  p 
a     o 


b       &  +  £ 


p-t 

Now approaches  zero  as  its  limit,  by  Theorem  VI. 

Therefore  —  approaches  -  as  its  limit. 


Example.      ^*2 


x 


x 


,^[-3]_-5    .5 

limit   r    _  K"l       -  7      7 


x 


im-2[*-5] 


As  illustrations  of  the  manner  of  applying  the  above  theo- 
rems, consider  the  following  examples  : 

Example  1.   Find    ™lt  \xn  |,  where  n  is  a  positive  integer. 
xn  =  x  •  x  '  x  •  ' '  •  to  n  factors. 

to  n  factors,  by  Theorem  V, 
=  a-a-a»"»  to  w  factors? 


LIMITS 


25 


Example  2.     Fiud  x±1 


■3aj*  +  a>-l" 


limit 
x 


X3  —  3  x  -\-  4 

J  [*-8.  +  4]-™  M  +  JS?  [-8«]  +  SSf  [4] 


limit 
by  Theorem  IV, 


=1-3+4 
=  2. 

Therefore,  since  the  limit  of  the  denominator  is  not  zero, 

3^  +  a,_1-|      ™!  [3a^  +  x-lj 


limit 
x=l 


x?-3x  +  4 


^itj-aj3_3a._i.4j 


,  by  Theorem  VII, 


Example  3.     Find 


x 

=  3 

2 

limit 


x  +  1' 


a2  +  3  a;  -  2 


Divide  both  numerator  and  denominator  by  x  raised  to  the 
highest  power  to  which  it  appears  in  the  fraction,  i.e.  by  x2. 


limit 

£  =  00 


r9/v.2 


•2s2-s  +  r 

_^  +  3  j,.  _  2_ 


__  limit 

X  =  VO 


~ 

1 

1~| 

2- 

-  + 

X 

x" 

1  + 

3_ 

2 



X 

JO 

=2. 


EXERCISES 

1.  Prove  that  if  the  numerator  of  a  fraction  does  not  ap- 
proach zero,  and  the  denominator  does  approach  zero,  the 
fraction  either  (1)  increases  without  limit,  or  (2)  decreases 
without  limit,  or  (3)  is  such  that  in  absolute  value  it  increases 
without  limit. 

2.  Prove  that  if  a  function  (1)  increases  without  limit, 
(2)  decreases  without  limit,  any  finite  multiple  of  the  function 
will  (1)  increase  without  limit,  (2)  decrease  without  limit. 

Choose  -J,  \ ,  f ,  £,  •••  as  a  succession  of  values  for  x,  and  for 
the  first  six  values,  plot  as  lengths  from  the  origin  on  a  straight 


26 


DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


line,  the  values  which  the  function  in  each  of  exercises  3  and  4 
assumes.  In  each  case,  what  is  the  length  of  the  region  deter- 
mined by  the  sixth  value  of  x,  within  which  the  extremities  of 
the  lengths  representing  the  function  must  lie  for  all  subsequent 
values  of  x? 


3. 


x 


4. 


3a  +  l 
6  ' 


x 


Choose  1,  2i,  If,  21,  •••  as  the  succession  of  values  for  a?,  and 
give  a  treatment  similar  to  that  in  exercises  3  and  4  for  the 
function  in  each  of  the  exercises  5  and  6. 


5. 


2a?  +  l 
x-3  ' 


6. 


x 


aj-f-1 


For  any  succession  of  values  for  x,  in  each  of  the  following 

a^  +  l" 


exercises,  find 


7. 


8. 


9. 


10. 


11. 


12. 


limit 
x=0 

limit 
x=l 

limit 
x=0 

limit 
x=l 

limit 
x±a 

limit 

X  =  <X) 


3" 
x 

3a2-2x  +  4' 
a?-2x  +  l 

(3a:-2)  (6a:  +  7)~ 
_(3x  +  4)  (a^-l) 

"(3  a;  -2)  (6^  +  7)" 
_(3a  +  4)  (z2-l)_ 

a?2  +  a#  +  a2"1 
a2-}-^2 

~  OS3  -f  SC2  +  1 


3  a?3  +  a?  —  4 


13. 


14. 


15. 


16. 


17. 


18. 


29.  Function  infinite  or  infinite  negatively  for  a  value  of  the 
variable.  Let  f(x)  be  a  single-valued  function  of  the  variable  x. 
The  function  f(x)  is  said  to  be  (1)  infinite,  (2)  infinite  nega- 
tively for  a  value  x0  of  x,  if  as  x  approaches  x0,  f(x)  becomes 
(1)  greater,  (2)  less  than  any  number  we  may  choose  to  name 


LIMITS 


27 


(see  Art.  22).  The  curve  y=f(x)  in  rectangular  coordinates, 
therefore,  will  be  (1)  infinite,  (2)  infinite  negatively  for  a  value 
x0  of  x,  if  as  x  approaches  x0  the  ordinate  of  the  curve  becomes 
and  remains  (1)  greater,  (2)  less  than  any  length  we  may  choose 
to  name. 

It  frequently  happens  that  a  function  is  infinite  or  infinite 
negatively  for  a  value  x0  as  x  approaches  xQ  being  always 
greater  than  it,  and  the  opposite,  namely  infinite  negatively  or 
infinite,  as  x  approaches  x0 
being  always  less  than  it. 


Thus,    the   function 


x 


x—2 

when  x  =  2,  is  infinite  if  x 

approaches   2   being   always 
greater  than  2,  and  is  infinite 
negatively   if   x  approaches 
2  being  always  less  than  2. 
In    the    neighborhood    of 


Y 

I 

0 

\ 

X 

Fig.  7. 


x  =  2,  the  curve  y  = 


x 


X 


is  as  in  Fig.  7. 


30.    Function  finite  between  two  values  of  the  variable.     Let 

f(x)  be  a  single-valued  function  of  x.  The  function  f(x)  is 
said  to  be  finite  between  two  values  xy  and  x2  of  x,  if  there 
is  no  value  x0  of  x  between  x1  and  x2,  for  which  fix)  does  not 
have  a  definite  value. 


Thus,  the  function 


x 


is  finite  between  any  two  values  of 


x-2 
x  which  do  not  contain  the  value  x  =  2  between  them. 


31.  Function  continuous  or  discontinuous  for  a  value  of  the 
variable.  Let  f(x)  be  a  single-valued  function  of  x.  Suppose 
that,  when  x  has  a  value  near  a  particular  value  x0,  f(x)  is  finite. 
Let  x0  —  h  and  x0  +  h  be  two  arbitrarily  chosen  values  of  x 
near  x0.     The  function  f(x)  is  said  to  be  continuous  for  x  =  x0, 


if 


limit 
h=0 


f(xQ-h) 


limit 
h=0 


f&o  +  h) 


=Kxo)- 


28  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

The  function  f(x)  is  said  to  be  discontinuous  for  x  =  x0,  if 


limit 
h=0 


f(xQ  -  h) 


=#= 


limit 
h=0 


f(x0  +  h) 


It  is  also  discontinuous  for  x  =  x0,  if 


limit 
h=0 


/(a*  -  K) 


limit 
h  =  0 


f(x0  +  h) 


} 


1^/w, 


but  this  case  is  of  rare  occurrence. 
In  the  curve  y  —  f(x)  in  rectan- 
gular coordinates,  x0  —  h,  x0  and 
Xq  +  7i  represent  abscissas,  and 
f(x0  -  h),  f(x0),  and  f(x0  +  ft), 
xr  ordinates.  In  Fig.  8,  f(x0  —  h)  is 
AB,  f(x0)  is  EG,  and  f(x0  +  h)  is 
CD.     If 


limit 
/i=0 


/(^o  -  K) 


limit 
h=0 


/(*o  +  &) 


=Xa;b)> 


which  is  the  case  in  this  figure,  the  curve  is  continuous  for 
x  =  x0,  or  at  the  point  E. 


0 


Fig.  9. 


Y 

F 

E 

I 

G^ 

D 

0 

A        E           ( 

■/ 

X 

Fig 

10. 

In  Figs.  9  and  10,  f(x0-h)  is  AB,  and  X^o  +  ft)  is  CD.     If 


limit 
/i  =  0 


X^o  -  h) 


L 


limit 
T  h  =  0 


JX?o  +  ft) 


which  is  the  case  in  either 


figure,  because  AB  approaches  the  limit  EG,  and  CD  the  limit 
EF,  the  curve  is  discontinuous  for  x  =  x0,  or  at  the  point  E. 

So  far  the  curve  was  supposed  to  remain  finite  for  values  of 
x  near  x  =  x0. 


LIMITS  29 

The  function  f(x)  is  also  said  to  be  discontinuous  for  a  value 
x  =  x0,  if  it  becomes  infinite  or  infinite  negatively  for  the  value 

X    &Q. 

32.  Definitions.  When  a  function  remains  finite  for  a  value  of 
the  variable  and  is  discontinuous  for  this  value,  it  is  said  to  have 
a  finite  discontinuity  for  this  value  of  the  variable. 

When  a  function  becomes  infinite  or  infinite  negatively  for  a 
value  of  the  variable,  it  is  said  to  have  an  infinite  discontinuity 
for  this  value  of  the  variable. 

33.  All  the  functions  with  which  the  student  is  likely  to  meet 
are  continuous  for  all  values  of  the  variable  excepting  those 
which  make  the  function  infinite  or  infinite  negatively. 

No  attempt  is  made  here  to  prove  a  function  continuous, 
and  it  will  be  assumed  in  the  following  chapters  that  all  the 
functions  considered  are  continuous  excepting  for  those  values 
of  the  variable  for  which  they  become  infinite  or  infinite 
negatively. 

EXERCISES 

1.  Plot  the  curve  y  =  -    x^~  '  n   for  values  of  x,  1st,  near 

(x—2)(x—3) 

1 ;  2d,  near  2 ;  3d,  near  3. 

x  4-  2 

2.  Plot  the  curve  y  =  - — ~w    for  values  of  x  near  1. 

(x  —  l)2 

3.  Show  that  x  (x  —  1)  is  continuous  when  x  =  l.  Plot  the 
curve  y=x(x  —  T)  for  values  of  x  near  1. 

4.  Is  the  function continuous,  1st,  when  x  =  1 ;  2d, 

1  +  e*  1 

when  x  =  0?    Plot  the  curve  y  = for  values  of  x  near 

these  values.  1  +  e* 

5.  Plot  the  curve  y  =  -sin#  for  values  of  x  near  0.     What 

x 

can  be  said  about  the  curve  when  x  =  0  ? 


CHAPTER   II 

DEEIVATIYES 

• 

34.  Definitions.  The  increase  in  a  variable  due  to  its  having 
passed  from  one  value  to  another  is  called  the  increment  of 
the  variable. 

The  increase  in  the  function  due  to  the  variable  having 
passed  from  one  value  to  another  is  called  the  increment  of 
the  function. 

An  increment  is  usually  denoted  by  writing  A  before  the 
variable,  A  being  merely  a  symbol  for  the  words  increment  of. 

For  example,  in  the  equation  y  =  x2,  suppose  that  the  variable 
has  passed  from  the  value  6  to  the  value  8.  Its  increment  is  2. 
That  is,  Ax  =  2.  When  x  passed  from  6  to  8,  y  passed  from 
36  to  64.  Its  increment  is  64-36,  or  28.  That  is,  Ay  =  28. 
Or  again,  suppose  that  x  has  passed  from  the  value  6  to  the 
value  4.  Its  increment  is  —  2.  That  is,  Ax  =  —  2.  When  x 
passed  from  6  to  4,  y  passed  from  36  to  16.  Its  increment  is 
16  -  36,  or  -  20.     That  is,  Ay  =  -  20. 

35.  In  the  equation  y  =  x*,  let  x  have  the  value  6,  and  calcu- 
late the  values  of  Ay  and  -^,  found  by  giving  Ax  certain 

Ax 

values.     The  values  given  to  Ax  and  the  values  resulting  for 
Ay  and  — ^  are  given  in  the  scheme  at  the  top  of  the  following 

L\00 

page.     From  a  study  of  this  scheme,  we  see  that,  in  this  par- 
ticular example,  as  Ax  approaches  zero,  Ay  approaches  zero, 

but  — ^  does  not  approach  zero.     It  approaches  12. 

Ax 


DERIVATIVES 


31 


If  Ax  = 

then  Ay  = 

-.  Aw 
and  ~  = 

Ax 

3 

45 

15 

2 

28 

14 

1 

13 

13 

.1 

1.21 

12.1 

.01 

.1201 

12.01 

.001 

.012001 

12.001 

.0001 

.00120001 

12.-0001 

Ax 

12  Ax  +  A? 

12  +  Ax 

36.  Instead- of  giving  x  the  value  6,  as  in  the  last  article,  we 
may  give  it  any  value  x0.  Indicate  by  y0  the  value  of  x2  when 
x  has  the  value  x0.  The  increment  of  y  is  found  by  giving  x 
the  value  x0  +  Ax,  where  Ax  is  an  arbitrary  value,  and  subtract- 
ing y0  from  the  result.     This  increment  by  definition  is  Ay. 


Then 

Divide  by  Ax. 
Pass  to  limits. 

This  gives  limitA 


Ay  =  (x0  +  Az)2  —  x02 

2 

=  2  XqAx  +  Ax  . 

—  =  2  cc0  +  Aoj. 

Aa? 


limit 
A:r  =  0 


Ax 


2xn. 


Ax. 


'Ay 
Ax 


limit 


if*»=6,rio 


'Ay 

Ax 


for  any  value  x0  of  a,*.     In  particular, 
=  12,  the  result  of  the  last  article. 


Since  x0  is  any  value  of  x,  we  may  drop  the  subscript  and 


say  that 


limit 
Ax  =  0 


Ay 

Ax 


=  2x  for  any  value  of  x. 


limit 


'Ay 
Ax 


,  when  x 


37.    Definition.     In  the  equation  y  =  f(x),  J^Vq 

has  some  particular  value  x0,  is  called  the  derivative  of  y  with 
respect  to  x  when  x  =  x0,  or,  sometimes,  the  differential  coeffi- 


cient of  y  with  respect  to  x  when  x  =  x0,  and  is  written 


dy 
dx 


32 


DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


Since  x0  is  any  value  of  x,  we  may  drop  the  subscript  and  say, 


limit 
A:c  =  0 

briefly. 


Ay 

Ax_ 
dy 
dx 


for  any  value  of  x  is 


dy 
dx 


This  is  written  more 


dy 


38.  In  an  equation  y  =  f(x)  the  successive  steps  in  finding  — 
are: 

1st.  Let  x  have  the  value  x0  and  calculate  the  corresponding 
value  yQ  of  y. 

2d.  Let  x  have  the  value  x0  +  A#,  where  A#  is  an  arbitrary 
increment,  and  calculate  the  corresponding  value  of  y.  This 
value  will  be  y0  +  the  increment  in  y  due  to  the  increment 
given  to  x,  or  y0  -f-  Ay. 

3d.     Subtract  y0  from  y0  +  A?/. 

4th.   Divide  Ay  by  Asc. 

5th.   Pass  to  limits. 

The  following  examples  will  illustrate  the  method : 


Example  1.   Find 
Let  x  =  5. 

Let  x  =  5  +  Ax. 
Subtract. 


dy 
dx 


when  y  =  Xs  —  2  x  +  4. 

z=5 

.-.  ?/0  +  A?/=(5  +  A^)3-2(5  +  A^)+4. 
.-.  A?/=(5  +  Aa;)3-2(5  +  Aa)+4 
-119 


=  73Atf  +  15A£2  +  Aa3. 


Divide  by  Ax. 
Pass  to  limits. 


.    ^2/_ 


limit 
Ax  =  0 


Ax 

'Ay' 
Ax 


=  73  +  15  Ax  +  Ax 


=  73  when  x  =  5. 


=  73. 


x=5 


DERIVATIVES 


33 


dy 


Example  2.    Find  l-^ 
dx 


when  y  =  a?  —  2  x  -f  4. 

Let  a;  =  x0.  .•.  y0  =  x03  —  2  x0  -f-  4. 

Let  a  =  x0  -f  Ax.  .-.  y0  +  Ay  =  (x0  +  Ax)3 —  2  (x0  4- Ax)  +  4. 

Subtract.       . •.  Ay  =  (x0 + Ax)3 — 2  (x0 + Ax) + 4  —  (x03—  2  x0 + 4) 
=  (3  xi  -  2)  Ax  +  3  x0Ax2  +  Ax3. 


Divide  by  Ax. 
Pass  to  limits. 


.-.  ^  =  3x02-2  +  3x0Ax  +  Ax2. 

Ax 


limit 
Ax  =  0 

dy 
dx 

dy 


Ay" 

Ax 


=  3  Xq  —  2  when  x  =  x0. 


_  3/Y.2 


3xn2-2. 


Example   3.     Eind  —  when  ?/  =  Vx. 

dx 


j-jeij  x  —  Xq. 
Let  x  =  x0  +  Ax. 
Subtract. 

Divide  by  Ax. 


.-.  ?/0  =  Vx0. 


2/o4-A2/  =  Vx0  +  Ax. 
.  •.  Ay  =  Vx0  +  Ax  —  Vx0. 


Ay  _  Vx0  +  Ax  —  Vx0 

Ax  Ax 


If  we  take  the  limit  of  this  expression  as  it  stands,  we 
encounter  the  indeterminate  form  -§-.  The  difficulty  can 
be  avoided  by  rationalizing  the  numerator.  Multiply  both 
numerator  and  denominator,  therefore,  by  Vx0  4-  Ax  4-  Vav 


.   4?/_ 


Ax 


Ax      Ax(  Vx0  +  Ax  +  Vx0)      Vx0  +  Ax  +  Vx0 

1 


Pass  to  limits. 


limit 

"  Ax  =  0 


Ay 
Ax 


2Vx0 

•  dy  _    1 

'  dx     2  Vx 


when  x  =  xn. 


34 


DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


EXERCISES 


1.    In  the  equation  y=x2—2x,  calculate  the  values  of  Ay  and 
— ^  corresponding  to  the  values  4,  3,  2,  1,  .1,  .01,  .001  of  Ax. 

AX  ,- 

Ay 

Ax 


when  a;  =  5.     Find  limitn 


when  x  =  5. 


2.    In  the  equation  y  =  xs,  calculate  the  values  of  A?/  and  — U. 

Ax 

corresponding  to  the  values  3,  2,  1,  .1,  .01,  .001  of  Ax,  when 

Ay 

Ax 


a;  =  10.     Find  limitA 


when  x  =  10. 


Find 


dy 


dx 


in  each  of  the  following  cases : 

3.  y  =  x2  —  2  x,  (a)  when  xQ  =  -J-;   (6)  when  #0  =  2. 

4.  y  =  (#  —  1)  (3  a;  +  4),  (a)  when  #0  =  3 ;  (5)  when  #0  =  4. 

5 .  y  =  xs  when  #0  =  3. 


6.    ?/  = 


x  +  a 


when  #ft  =  a. 


dv 
Find  -^  in  each  of  the  following  cases  : 
dx 


7.    y  =  x2. 


8.   y  = 


9.    2/  =  V#. 

cc  +  1 


Va  +  i 


10.    2/ 


x2  +  l 


x  +  2 
*      a2+l 


CHAPTER   III 

GEOMETRICAL  INTERPRETATION  OP  A  DERIVATIVE. 
PROBLEMS  IN  SPEED 

39.  We  shall  now  consider  the  geometrical  interpretation  of 
a  derivative. 

Let  y  =f(x)  be  the  equation  of  a  given  curve.  To  avoid 
complications,  we  shall  suppose  that  f(x)  is  single  valued  and 
continuous.  Since  f(x)  is  not  known,  we  cannot  plot  the 
curve,  but,  as  we  wish  a  geometrical  picture  of  the  equation, 
we  shall  suppose  that  part  of  the  curve  at  least,  if  it  could  be 
plotted,  would  be  as  drawn  in  Fig.  11.  The  student  can 
readily  satisfy  himself  that  the  same  reasoning  will  hold  for 
the  curve  drawn  in  any  other  position. 


Y 

A 

C 

s/$              /  r 

'                /     0 

X 

Fig.  11. 


To   let   x  assume  a  particular  value  x0  in  finding 


dy 
dx 


is 


geometrically  to  choose  a  length  measured  along  the  ic-axis  as 
the  abscissa  of  a  point  on  the  curve.  To  calculate  the  corre- 
sponding value  of  y  is  geometrically  to  find  the  corresponding 
ordinate.     Then  the  coordinates  of  A  are  (x0,  y0).     Similarly, 

35 


36 


DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


the  coordinates  of  B  are  (x0  +  Ax,  y0  +  Ay) .  The  result  when 
2/o  is  subtracted  from  y0  +  A?/  is  A?/.     In  the  figure,  Ay  is  CI?. 

Now  — ^  =  tan  <£,  where  <£  =Z  (TAB.     Therefore  — "  is  the  slope 

Ax  Ax 

of  the  secant  line  through  the  points  A  and  B.  As  A#  =  0, 
_B  =  A  and  the  points  of  intersection  of  the  secant  line  approach 
coincidence.  The  limit  of  the  secant  line  as  B  approaches  A 
is  by  definition  the  tangent  line  to  the  curve  at  the  point  A. 
Let  the  angle  which  the  tangent  line  at  the  point  A  makes 
with  the  avaxis  be  denoted  by  t.  As  Ax  approaches  zero,  cj> 
approaches  r  and 

=  tan  r. 


limit 
A;r  =  0 


Ay 

Ax 


Therefore 


dy 
dx 


is  geometrically  the  slope  of  the  tangent  line 


x=x0 


to  the  curve  y=f(x)  at  the  point  (x0,  y0)  on  the  curve. 


40.   As  an  illustration  consider  the  geometrical  interpretation 

of  ^ 

dx 


in  the  parabola  y2  =  4  x. 

To  put  the  equation  in  the  form  y=f(x),  solve  for  y. 
.".  y  =  ±  2  -y/x.  This  is  a  double-valued  function  of  x  (see 
Art.  14).  We  shall  here  consider  only  y  =  -\-2-\/x  and  con- 
fine our  attention  to  the  part  of  the  curve  above  the  #-axis. 

-LieTJ     x  —  Xq. 

Let    x=x0-\-Ax. 


r.y0+Ay=2Vx0+Ax. 

Subtract. 

.-.  Ay =2  V#o  -\-Ax—2 V&v 

Divide  by  Ax. 

Ay  _  2  Vff0+ Ax—  2  V^0_ 
Ax  Ax 


GEOMETRICAL   INTERPRETATION   OF  DERIVATIVE        37 

...  ^l=  JL -(see Example 3, Art. 38). 

Aa;      Va^o  +  Ace  +  VxQ 


tan  r  =  — 

de- 


limit 
Ax  =  0 


"A.ty" 


Xo 


Therefore  the  slope  of  the  tangent  line  to  the  curve 
y  =  +  2  V&*  or  the  part  of  the  parabola  y2  =  4  x  above  the  x-axis, 
at  any  point  (x0,  y0)  on  the  curve,  is  — —••  In  particular,  if 
x0  =  1,  tan  T  =  1,  or  t  =  45°.  ^xo 

EXERCISES 

1.  In  the  curve  y  =  —  2-y/x,  find  the  slope  of  the  tangent 

line  at  any  point  (xQ,  y0)  on  the  curve.  Ans. — 

Vfl?0 

2.  What  is  the  angle  which  the  tangent  line  to  the  parabola 

y2  =  4  x  makes  with  the  a>axis  at  the  origin  ?  How  does  the 
angle  change  as  the  point  (aj0,  y0)  on  the  parabola  is  made  to 
move  farther  out  on  the  part  of  the  curve  above  the  sc-axis  ? 
On  the  part  below  the  a>axis  ? 

3.  Find  the  angle  which  the  tangent  line  to  the  curve  y  =  cc3 
makes  with  the  cc-axis  at  the  point  (—  1,  —1);  at  the  point 
(0,  0) ;  at  the  point  (2,  8).  Ans.  tan"1 3,  0,  tan"1 12. 

Given  that  the  equations  of  the  tangent  and  normal  lines  to 
the  curve  y=f(x)  at  a  point  (x0,  y0)  on  the  curve  are 


dy 

y-y<>  =  -r 

ax 


dx 

find: 


(x  —  x0),     equation  of  the  tangent  line, 
(x  —  x0),     equation  of  the  normal  line, 


X=Xq 


4.    The  equations  of  the  tangent  and  normal  lines  to  the 

parabola  x2  =  4  y,  at  the  point  (2,  1)  ;  also  at  the  point  (—  2, 1). 

x  —  y  =  1. 
Ans.  y 

x  +  y  =  —  l. 


38         DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

5.  The  equations  of  the  tangent  and  normal  lines  to  the 
ellipse   4  x2  +  y2  =  8,  at  the   point   (1,  2)  ;   also   at   the   point 

(~1>2)'  A       2x  +  y  =  4:. 

Ans.  ■  J 

2x-y  =  -4:. 

41.  Definitions.  The  mean  or  average  speed  of  a  moving 
body  is  the  distance  passed  over  by  the  body  in  a  given  time, 
divided  by  the  time,  the  distance  being  expressed  in  units  of 
length  and  the  time  in  units  of  time. 

For  example,  if  a  body  passes  over  60  feet  in  2  seconds,  its 
mean  or  average  speed  is  30  feet  per  second. 

The  actual  speed  of  a  moving  body  at  a  given  instant  is  the 
limit  which  the  mean  speed  for  a  period  of  time  immediately 
succeeding  the  instant  in  question  approaches  as  the  length  of 
the  period  of  time  is  allowed  to  become  indefinitely  decreased. 

The  mean  angular  speed  of  a  body  rotating  in  a  plane  about 
a  point  is  the  angle  through  which  the  body  has  turned  in  a 
given  time  divided  by  the  time,  the  angle  being  expressed  in 
units  of  angle  and  the  time  in  units  of  time. 

For  example,  if  a  body  rotates  through  an  angle  of  60°  in 
10  seconds,  its  mean  or  average  angular  speed  is  6°  per  second. 

The  actual  angular  speed  of  a  body,  rotating  in  a  plane 
about  a  point,  at  any  instant,  is  the  limit  which  the  mean  speed 
for  a  period  of  time  immediately  succeeding  the  instant  in 
question  approaches  as  the  length  of  the  period  of  time  is 
allowed  to  become  indefinitely  decreased. 

When  there  is  danger  of  confusion  between  speed  along  a 
curve  and  angular  speed,  the  former  is  called  linear  speed. 

The  speed  of  a  moving  body  is  said  to  be  uniform  when  the 
actual  speed  at  £very  instant  is  the  same. 

The  speed  of  a  moving  body  is  said  to  be  variable  when  the 
actual  speed  is  continually  changing. 

Actual  speed  is  usually  spoken  of  simply  as  speed. 

Another  name  for  speed  is  rate  of  motion,  or  more  simply 
rate. 


PROBLEMS  IN  SPEED  39 

42.  Units  of  speed.  The  measure  of  mean  linear  speed  is 
that  of  a  certain  distance  divided  by  that  of  a  certain  time. 
The  unit  of  mean  linear  speed  is  therefore  the  speed  of  a 
certain  point  which  moves  through  a  unit  of  distance  in  a 
unit  of  time.  Expressed  in  English  units  it  may  be  1  foot 
per  second,  1  mile  per  hour,  etc.,  and  in  French  units  1  centi- 
meter per  second,  1  kilometer  per  hour,  etc.  The  measure  of 
a  mean  angular  speed  is  that  of  a  certain  angle  divided  by 
that  of  a  certain  time.  The  unit  of  angle  usually  employed 
is  the  radian.  The  unit  of  mean  angular  speed  in  terms  of 
the  radian  is  therefore  1  radian  per  unit  of  time. 

Since  the  actual  speed  of  a  body  at  a  point  is  expressed  in 
terms  of  the  limit  of  mean  speed,  its  unit  is  that  of  the  mean 
speed. 

43.  To  find  the  speed  of  a  body  at  any  particular  instant, 
we  must  find  the  mean  speed  of  the  body  for  a  period  of  time 
immediately  succeeding  the  instant  in  question,  and  then  find 
the  limit  which  this  mean  speed  approaches  as  the  period  of 
time  is  allowed  to  become  indefinitely  decreased. 

As  an  example  in  finding  the  speed  of  a  moving  body  at  any 
particular  instant,  consider  the  following  problem : 

A  body  falls  freely  from  rest  in  a  vacuum.     Eind  its  speed 

at  the  end  of  t0  seconds. 

By  experiment,  it  has  been  found  that  the  distance  in  feet 

passed  over  by  a  body  falling  freely  from  rest  in  a  vacuum 

during  a  time  t  in  seconds  is  s  =  16 12. 

Then,  in  the  equation    s  =  16  f, 

let  t  =  t0.  .-.  s0  =  16 102  =  distance  in  feet  the  body 

has  fallen  in  t0  seconds. 

Let  t=t0+&t.    .-.  s0  -j-  As  =  16  (t0  +  &t)2  =  distance  in 

feet  the  body  has  fallen 

■  -A  in  t0  +  A£  seconds. 

As  .-.  As=16(£0+A£)2-16£02=distance  in  feet  the  body 

has  fallen  in  At  seconds 
Fig.  13. 

from  point  A.     (Fig.  13.) 


40  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

As 

.*.  —  =  32t0  +  16  At  =  mean  speed  of  the  body  over  the  dis- 
tance As,  expressed  in  feet  per  second. 


ds 
dt 


limit 
At=0 


"As" 

At 


32  t0  =  actual  speed  of  the  body  at 


the  point  A,  or  at  the  end  of  the  time  t0,  expressed  in  feet  per 
second. 

Note.  The  distance  in  feet  passed  over  by  a  body  falling  freely  from 
rest  in  a  vacuum  during  a  time  t  in  seconds  is  s  =  a  gt2,  where 

g  =  32.0894(1  +  0.0052375  sin  I)  (1  -  0.0000000957  e),* 

I  being  the  latitude  of  the  place  and  e  the  height  in  feet  above  the  sea 
level.  This  gives  g  the  value  32.2  approximately.  In  this  problem  it 
was  chosen  32. 

44.  Definition.  The  equation  by  which  the  distance  s  passed 
over  by  a  moving  body  in  a  time  t  is  expressed  as  a  function 
of  t,  is  called  the  law  of  motion  of  the  body. 

Thus,  in  the  example  of  the  preceding  article  the  distance  s 
passed  over  by  a  body  falling  freely  from  rest  in  a  vacuum 
during  a  time  t  is  s  =  -J-  gt2.  Then  s  =  \  gt2  is  the  law  of  motion 
of  this  body. 

In  all  cases  the  law  of  motion  of  the  body  must  be  deter- 
mined before  the  speed  of  the  body  at  any  given  instant  can 
be  found.  This  law  is  always  derived  with  more  or  less 
mathematical  manipulation  from  observed  measurements. 
Thus,  the  law  of  the  example  of  Art.  43  was  determined  by 
experiment. 

45.  As  an  example  in  which  mathematical  manipulation 
must  be  combined  with  observed  measurements  in  order  to 
determine  the  given  law,  consider  the  following  problem  : 

One  end  of  a  ladder  20  feet  long  rests  on  the  ground,  12 
feet  from  the  foundation  of  a  building.  The  other  rests 
against  the  side  of  the  building.  If  the  end  on  the  ground  is 
being  carried  away  from  the  building  on  a  line  perpendicular 

*  See  Maurer's  Technical  Mechanics,  Part  I,  Art.  13. 


PROBLEMS  IN  SPEED 


41 


to  it  at  the  uniform  rate  of  4  feet  per  second,  find  the  law 
of  motion  of  the  other  end. 

Here,  the  observed  measurements  are  the  rate  at  which  the 
end  of  the  ladder  on  the  ground  is  moving,  the  length  of 
the  ladder,  and  the  distance  of  the  foot  of  the  ladder  from 
the  building  when  in  its  initial  position.  To  determine  the 
law  from  these  measurements : 

Let  AB  (Fig.  14)  represent  the 
side  of  the  building,  and  BG  the 
line  perpendicular  to  it.  Let  t 
represent  the  number  of  seconds 
during  which  the  ladder  is  moving. 
Then  12  +  4 1  is  the  distance  of 
its  foot  from  the  building  at  the 
end  of  the  time  t.  Let  s  represent 
the  distance  through  which  the  top 

of  the  ladder  has  moved.  If  D  is  the  point  at  which  the 
ladder  rested  in  its  initial  position,  and  F  the  point  it  has 
reached  at  the  end  of  t  seconds,  then 

s  =  DF 
=  DB-FB 


Fig.  14. 


=  V400  -  144  -  V400  -  (12  +  4  ff 


=  lQ-4;-y/l6-6t-t2. 


Therefore  s  —  16  —  4 Vl6  —  6 1  —  f  is  the  law  of  motion  of 
the  top  of  the  ladder. 

46.  If  the  law  of  motion  of  a  body  can  be  formulated  so 
as  to  express  the  distance  s  passed  over  by  the  body  as  a 
function  of  the  time  t,  then  the  speed  of  the  body  at  the 


t=tn 


can  be  determined  as  in  the  example 


els 
time  t,  namely  — 

of  Art.  43.  dt 

As  a  further  illustration  of  finding  the  speed  of  a  body  at 

a  given  instant  when  the  law  of  motion  is  known,  find   the 


42         DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

speed  at  which  the  top  of  the  ladder  in  the  problem  of  the 
preceding  article  is  moving  at  the  end  of  (a)  1  second; 
(b)  1\  seconds. 


To  find  these  speeds,  find  — 
in  turn  in  the  result. 


,  and  substitute  1  and  1\ 


Let  t  =  t0.      .-.  s0  =  16  -  WlQ- 6  t0-t02. 


Let  t  =  t0  +  At.     .-.  s0+As=16-Wl6-6(t0+At)-(t0  +  Aty 


As^      ±  Vl6  -  6ft  +  AQ  -  fa  +  At)2 -  Vl6  -  6 10 -  t£ 

"At      '  At 


Eationalize  the  numerator  and  pass  to  limits. 

ds 

"  dt 


=  1  ^o  +  3 


ds 
dt 

ds 
dt 


t=t0         Vl6  —  6t0  —  t02 

=  4  — =  =  —-  feet  per  second,  and 
t=i  V9      3 

=  4  — *_  =  |4 Vl9  feet  per  second. 


EXERCISES 

1.  A  man  6  feet  high  walks  directly  away  from  a  lamp- 
post 10  feet  high  at  the  uniform  rate  of  4  miles  per  hour. 
How  fast  does  the  end  of  his  shadow  move  ?  Does  it  move 
uniformly  ?  Ans.  10  mi.  per  hr.     Yes. 

2.  In  Exercise  1,  how  fast  is  his  shadow  lengthening? 
Does  it  lengthen  uniformly  ?  Ans.    6  mi.  per  hr.     Yes. 

3.  A  man  is  walking  up  a  plane  inclined  45°  to  the  ground 
at  the  rate  of  2  miles  per  hour.  At  what  rate  is  his  projection 
on  the  ground  moving  ?  Ans.    V2  mi.  per  hr. 

4.  At  what  rate  must  a  man  walk  up  a  plane  inclined  60° 
to  the  ground  in  order  that  his  projection  on  a  plane  per- 
pendicular to  the  ground  may  move  at  the  rate  of  1  mile 

per  hour  ?                                                         .          2  , 

1  Ans.    mi.  per  hr. 

V3 


PROBLEMS  IN   SPEED  43 

5.  The  radius  of  a  spherical  soap  bubble  is  increasing 
uniformly  at  the  rate  of  ^  inch  per  second.  Find  the 
rate   at  which  the  volume  is   increasing  when  the  diameter 

is  -^y-  inch.  Ans.    —£—  cu.  in.  per  sec. 

4000 

If  a  body  with  an  initial  speed  v0  moves  down  a  smooth 
plane  inclined  «°  to  the  horizontal,  the  law  of  motion  of  the 

body  is  s  =  %gt28ma  +  v0t. 

Assuming  this  law,  solve  the  following  problems,     (g  =  32.) 

6.  If  a  body  starts  from  rest  and  moves  down  a  smooth 
plane  inclined  60°  to  the  horizontal,  find  its  speed  1  second 
after  it  starts.  Ans.   16V3  ft.  per  sec. 

7.  If  a  body  with  an  initial  speed  of  10  feet  per  second 
moves  down  a  smooth  plane  inclined  30°  to  the  horizontal, 
find  its  speed  when  it  has  gone  10  yards,  and  its  mean  speed 
for  this  distance.         Ans.   32.56  ft.  per  sec. ;  21.28  ft.  per  sec. 

47.  So  far  we  have  considered  the  speed  of  a  body  when  the 
law  of  motion  can  be  formulated  so  as  to  express  the  distance 
passed  over  as  a  function  of  the  time.  We  shall  now  consider 
a  slightly  more  general  problem,  one  in  which  the  ratio  of  the 
speeds  of  two  bodies  can  be  determined  when  the  law  of  motion 
can  be  formulated  so  as  to  express  the  distance  passed  over  by 
one  as  a  function  of  that  passed  over  by  the  other  in  the  same 
time. 

48.  As  an  example,  consider  the  following  problem: 

Suppose  that  two  bodies  are  moving  in  such  a  manner  that 
in  a  time  t  one  passes  over  a  distance  y  and  the  other  over  a 
distance  x,  where  y  =  x2  +  3  x  -f-  2.  To  find  the  ratio  of  the 
speeds  of  the  two  bodies  at  the  end  of  a  time  t0. 

At  the  end  of  a  time  t0,  one  has  passed  over  a  distance  x0  and 
the  other  over  a  distance  y0,  where 

y0  =  %o2  +  3x0  +  2. 


44 


DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


At  the  end  of  a  time  tQ  +  At,  one  has  passed  over  a  distance 
Xq  +  Ax  and  the  other  over  a  distance  y0  +  Ay,  where 

2/o  +  Ay  =  (a?0  +  Aa)2  +  3  (a0  +  Aa)  +  2. 

.-.  A#  =  (a0  +  Aa)2  +  3(a0  +  Aa)+2-(a02  +  3a0  +  2) 

=  (2a0  +  3)Aa  +  Aa2. 

Ay  _  (2  a0  +  3)  Aa  +  Aa2 
A£  At 

=  (2a0  +  3)Aa  +  Aa2  <  Aa 

Aa  A£? 

by  dividing  and  multiplying  by  Ax. 
2-1 


limit 
AJ=0 


'Ay 

At 


limit 
At=0 


"(2  x0  +  3)  Aa  +  Aa 

Ax 


Now  as  A£  =  0,  Aa;  =  0. 


limit 
At=0 


Ay 

At 


_  limit 
Ax  =  0 


"(2a0  +  3)Aa  +  Aa 

Ax 


2-1 


limit 
A£  =  0 


limit 
A«=0 


Ace 
A* 


"Aa" 
~At 


.    dy 
dt 


=  (2a0  +  3) 


dx 
dt 


Now 


eZa 
"eft 


and  — 
dt 


are  the  speeds  at  the  end  of  a  time 


t0  of  the  bodies  that  move  over  distances  x  and  y  respectively 
in  a  time  t.  Therefore  the  ratio  of  the  speeds  of  the  two 
bodies  is  such  that 


dy 

dt 


t=tn 


dx 

dt 


2a0  +  3. 


It  will  be  noticed  that  2  x0  +  3  is 


example, 


dy 
dx 


.    Therefore,  in  this 


dy 
dx 


dy 
dt 
dx 
dt 


PROBLEMS  IN   SPEED 


45 


49.  It  can  be  readily  proved  in  general  that  if  the  distances 
passed  over  by  two  bodies  in  a  time  t  are  x  and  y  respectively, 
and  y  can  be  expressed  as  a  function  of  x,  then 


dy 
dx 


dy 

dt 


dx 

dt 


where  x0  is  the  value  of  x  when  t  —  t0. 
Proof.     For  any  value  tQ  of  t, 


—  =  —  •  — ,  identically. 

At      Ax     At  J 


.     limit 
'  '   A£=0 


'Ay 

~At 


Now  as  At  =  0,  Ax  =  0. 


.     limit 


~Ay 

~At, 


limit 

At  =  0 


limit 
Ax=0 


'Ay 

Ax 


'Ay 
Ax 


limit 
A£  =  0 


limit 
At=0 


'Ax' 
~At 


'Ax' 


dy 
"  ~dt 


dy 
dx 


t=tr 


dy 
dx 


dx 

dt 


dy 

dt 

t=t0 

dx 

dt 

t=t0 

50.  In  the  curve  y  =f(x),  x  is  the  abscissa  and  y  the  ordinate 
of  any  point  (x,  y)  on  the  curve.  If  Ave  look  on  the  curve 
as  being  generated  by  a  moving  point  whose  coordinates  are 


(x,  y),  then 


dx 
dt 


and 


dy 

dt 


are  the  rates  at  which  the 


t  =t„ 


point  is  moving  parallel  to  the  x  and  y  axes  respectively  at 
the  end  of  a  time  t0.     In  other  words, 


dx 

dt 


and 


dy 

~dt 


are 


t=tn 


the   rates   at   which  the   abscissa   and   ordinate   respectively 


46         DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


are  changing  in  length  at  the  end  of  the  time  t0.     Therefore 

—  is  the  ratio  of  the  rate  of  change  in  length  of  the 

(XX  ~ 

x~  xo 

ordinate  to  that  of  the  abscissa  of  the  curve  at  a  point  (x0,  y0). 
As  seen  in  Art.  39,  it  is  also  the  slope  of  the  tangent  line 
to  the  curve  at  this  point. 

EXERCISES 

1.  If  y  =  3  or2  — 2  £  +  4,  find  the  ratio  of  the  rate  of  change 
of  y  to  that  of  x  when  x  =  2.  Ans.  10. 

2.  In  y  =  x2  +  3x  —  2,  what  is  the  value  of  x  at  the  point 
where  y  increases  6  times  as  fast  as  x  ?  Ans.  x  =  -§-, 

3.  For   what   values   of   x   do    y  =  xs  —  3  x2  +  2  x  —  2   and 

y  =  x2  —  6  change  at  the  same  rate  ?  4  +  VlO 

Ans.  x  =  — . 

3 

4.  Find  the  coordinates  of  the  point  at  which  y=3x2-\-2x— 2 
changes  at  the  same  rate  as  the  slope  of  the  tangent  line  to 
the  curve  that  represents  this  equation,  at  the  point. 

Ans.  (|,  |). 

Water  flows  from  a  vessel,  in  the  form  of  a  right  circular 
cylinder  of  radius  1  foot,  into  one  of  the  form  of  an  inverted 
circular  cone  of  semivertical  angle  30° : 

5.  If  the  level  of  the  water  in  the  cylinder  falls  uniformly 
at  the  rate  of  2  inches  per  minute,  at  what  rate  is  the  water 
flowing  ?  Ans.  288  it  cu.  in.  per  min. 

6.  With  the  above  rate  of  flow,  at  what  rate  will  the  level 
of  the  water  in  the  cone  be  rising  when  the  depth  is  6  inches  ? 
When  1  foot  ?     When  2  feet  ?  Ans.  24,  6,  1.5  in.  per  min. 

7.  At  what  depth  of  water  in  the  cone  will  the  level  be 
rising  just  as  fast  as  it  is  falling  in  the  cylinder  ?  At  what 
depth,  three  times  as  fast  ?  Ans.  20.78  in. ;  12  in. 


CHAPTER   IV 

GENERAL  FORMULAS  FOR  DIFFERENTIATION 

51.  Definition.  The  process  of  finding  the  derivative  of  a 
function  is  called  differentiation. 

In  all  cases  differentiation  may  be  performed  by  following 
the  steps  mentioned  in  Art.  38.  It  is,  however,  inconvenient 
to  be  compelled  to  resort  to  this  method  in  each  particular 
problem,  so  we  shall  derive  rules  or  formulas,  called  formulas 
for  differentiation,  by  the  aid  of  which,  labor  may  be  avoided 
in  finding  the  derivative  of  a  function.  In  the  next  article 
seven  such  formulas  are  derived.  They  are  numbered  from 
I  to  VII  for  convenience  of  reference.  As  will  be  seen,  they 
apply  to  all  the  classes  of  functions  given  in  Art.  7.  Hence 
the  name  "  General  Formulas." 

52.  GENERAL  FORMULAS   FOR   DIFFERENTIATION 


I. 

dx  _ 
dx 

=  1. 

II. 

dc  _ 
dx 

=  0. 

III. 

d( 

u  +  v)  _ 

_  du      dv 

dx 

(J.JL>         \Aj\ju 

IV. 

d(uv) 

dv  ,     du 

-u 1-  v — 

dx 

(A/Jb                 (JjJU 

v. 

d(cu)  _ 

du 

-  G 

dx 

dx 

d(-) 

du        dv 

V u 

Vol 

LtJU                        (jjtJO 

VI. 

\     / 

dx 

V2 

VII. 

dun  _ 

m_l  OjU 

=  nun  y — • 

dx 

dx 

47 


48 


DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


In  these  formulas,  u  and  v  are  functions  of  the  independent 
variable  x ;  c  is  any  constant ;  and  n  is  any  rational  constant. 

Note.     Formula  VII  is  true  for  n  auy  constant,  but  we  shall  consider 
it  only  in  the  cases  where  n  is  a  rational  constant. 


53.    Derivation  of  formulas. 

Proof  of  I. 

Let 

y  =  x. 

J-iCu  X  ^=  Xy. 

•'•    Vo  =  X0' 

Let  x  =  x0  -f-  Ax. 

.:  y0  +  Ay  =  x0  +  Ax. 

Subtract. 

.'.  Ay  =  Ax. 

Divide  by  Ax. 

'  '  Ax 

Pass  to  limits. 


dy 
dx 


limit 

Ax  =  0 


Ay 

Ax 


=  1. 


dy  _ 


dx 


=  1,  or  since  y  =  x,  — =  1. 
dx  dx 


Proof  of  II. 

Let 

y  =  c. 

-LiGXi  X  —  Xy. 

.-.  y0  =  c. 

Let  x  =  x0  +  Asc. 

.:  y0  +  Ay  =  c. 

Subtract. 

.-.  Ay  =  0. 

Divide  by  Ax.  m 

Ax 

Pass  to  limits. 

.   fy\        __  limit 
/7/v.l             Ax  =  0 

"^n=o. 

dy      A  •  dc      n 

.'.  —  =  0,  or  since  2/  =  c,  —  =  0. 


GENERAL  FORMULAS  FOR  DIFFERENTIATION      49 


Proof  of  III. 

Let 

Let  x  =  .To- 
Let  x  =  t0  +  At. 
Subtract. 

Divide  by  Ax. 
Pass  to  limits. 


y  =  u  +  V. 

.-.  yQ  =  uQ  +  v0. 
yQ  +  Aw  =  u0  +  Aw  +  w0  +  Aw. 

.-.  Aw  =  Azt  +  Aw. 

Aw  _  Aw      Aw 

At      At      At 


cZt 


limit 
Ax  =  0 


limit 
Ax  =  0 


"Aii      Aw" 
At      At 


c?w 

C?T 


AwT 

At 

+ 


+ 


limit 
Ax=0 


"Aw" 
At 


dv 


dy      du  ,  dv 

-?-  = 1 ,  or  since  y  =  u  +  v, 

CvVO  (J/iAj  Ltth 


dx 

d(u-\~v)  _  du      dv 


dx 


dx      dx 


A  similar  proof  will  apply  to  any  number  of  functions,  so 

"that  _,     •  .        ,         _        , 

a{u -\-v  +  w-\-  •••) _au  ,  dv  ,  dio  , 


dx                   dx     dx     dx 

Proof  of  IV. 

Let 

y  =  uv. 

J_jGTj    T Trv. 

.'.  w  =  w0w0. 

Let  t  =  t0  +  At. 

•*•  Vo  +  Aw  =  (w0  +  Aw)(w0  +  Aw). 

Subtract. 

•  '•  Aw  =  (w0  +  Aw)  (w0  +  Aw)  —  w0w0 

=  w0Aw  -f-  w0Aw  +  AwAw. 

Divide  by  At. 


A?/         Aw  ,       Aw  .    A     Aw 
.*.  -^  =  w0— -f  w0— +  Aw  — ■ 

At         At         At  At 


50         DIFFERENTIAL  AND  INTEGRAL   CALCULUS 
Pass  to  limits. 


dx 


limit 

Ax  = 


ait  fAy"] 
=°|_AaJ 


limit 


Aw 


,         Al6    . 


Ai>~ 


Ax=0|     °Aa;  •     -Aa.  ■         Aa. 


limit 
Ax=0 


Mr 


AiT 
Ace 


limit 
Ax=0 

limit 
Ax=0 


All 

A& 


Now   Alfa?* 

Ax  =  0 


limit 
Ax  =  0 


M< 


and    Alimitn 


Aw 


Av" 
'Ax 

Aw" 

Aft 

Av" 
Ait 


=  v, 


limit 
Ax=0 

limit 

°Ax=0 


Av 

Ax 

'Au 
Ax 


=  u( 


dv 
dx 


Au 


Av 
Ax 


du 

dx 


d(itv) 
dx 


limit 
Ax  =  0 


dv 

=  w0  — 

dx 


Au 


limit 
Ax  =  0 


Av 
Ax 


=  0- 


dv 
dx 


=0. 


du 
dx 


dx 
Proof  of  V. 

Let 

_Li6Tj  X  ^=  Xq. 

Let  x  =  x0  +  Ace. 
Subtract. 

Divide  by  Ace. 
Pass  to  limits. 


d(uv)         dv  .     du 
dx        dx 


y  =  cu. 
.\  yQ  =  cu0. 

.*.  Ay  =  cAu. 

A?/        Au 
.-.  — -  =  c 


dy 
dx 


Ax        Ax 

.    limit 
Ax  =  0 


'Ay 

Ax 


'~|        limit  Tc  Aw] 
•J       Ax=o|^  Aa.J 

_  c  limit  [~Am~] 
Ax=0^Aa.J 

(fat 


=  C 


d(cu)  _    du 
dx  dx 


dx 


GENERAL   FORMULAS  FOR  DIFFERENTIATION       51 


Proof  of  VI. 

Let 

Let  x  =  x0. 
Let  x  =  x0  +  Ax. 
Subtract. 


Proof  of  VII. 

Let 

At  first  suppose 

_ijeij  x  —  Xq. 

Let  x  =  xQ  +  Ax. 

Subtract. 


u 

V 


u0 

.-•  2/0  +  &y 


v0 

u0  +  Aw 


v0  +  Av 


•    Ay  =  U° +  A^      ^° 
v0  +  Av 


Divide  by  Ax. 


Pass  to  limits.     .*.  — 


Vn 


_  v0Au  —  UqAv 
v0(v0  +  Av) 

Au         Av 
A?/  _      A&*  Ax 

Ax~  v0(v0  +  Ao) 


dy 
dx 


_  limit  l~&y~ 
^x=^[_Ax 


limit 
Ax  =  0 


^o 


limit 


"Ait" 
Ax 


u 


Au         Av 
v u0-— 

Ax         Ax 

v0(v0  +  Av)  . 
limit 


°Ax  =  0 


Av 
Ax 


*££>[«.+*>] 


Vn 


du 

ax  x-x 


—  Uc 


dv\ 


KAjtAj  y—y, 


du 

d(  -  )      v u 

dx 


dv 
dx 


dx  v2 

y  =  un. 
that  n  is  a  positive  integer. 
.-.  yQ  =  uQn. 
.-.  y0  +  Ay  =  (u0  +  Au)n. 

•••  Ay=(u0  +  Au)n-u0n. 


52 


DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


Now  (w0  4-  Aw)w  —  u0n  can  be  factored  like  an  —  w0w.      Let 
w0  +  Aw  =  a. 

The  factors  are  a  —  w0  and  (a71'1  +  an~2u0+an~3Uo2-\ \-au0n~2 

+  wo71"1)- 

Substitute  w0  +  Aw  for  a  in  these  factors. 

.-.  (w0  +  Aw)ra  —  w0w  =  Aw  j  (w0  +  Am)""1  +  w0(w0  +  Aw)n~2 
+  w02(w0  +  Aw)w~3  +  -.  +  Wo""2(wo  +  Aw)  +  ^cT1  j  • 
Divide  by  Ax. 

.*•  f^  =  ~  \  Oo  +  ^)n"1+  ^o("o  +  Aw)-2 

Ax      Ax 


Pass  to  limits. 


+  w02(w0  +  Aw)w-3+...  +  ^ow-1S 


dx 


limit 
Ax=0 


,  limit 
Ax  =  0 


A?/ 

Ax 

Aw 

Ax 


;(^o+Aw)w-1+w0(w0+Aw)w-2+w02(w0+A?6^ 
As  Ax  approaches  zero,  Aw  approaches  zero,  and 


dy 


du 


(XX   x=xn        (XX 


\  W*-1  +  w^-1  +  Wow-x  +  •  •  •  +  Wo""1 1 , 

there  being  n  terms  in  the  parenthesis, 


n 1  (XXI 

=  nu0n  L  — 
dx 

...  —  =  www_1 — ,  when  n  is  a  positive  integer. 

(XX  (XX 

Next,  suppose  that  n  is  a  positive  fraction  =  -,  where  p  and 

q  are  positive  integers. 

z 
Then  y  =  uq. 

Raise  both  sides  of  the  equation  to  the  gth  power. 

...    y«  =  up. 

dy9  _dup 
dx       dx 


GENERAL  FORMULAS  FOR  DIFFERENTIATION      53 

Now  up  is  of  the  form  un  where  n  is  a  positive  integer, 
because  p  is  a  positive  integer. 

dup  _      p_!  ch^ 
da;  cto 

Also,  yq  is  of  the  form  un  where  n  is  a  positive  integer, 
because  g  is  a  positive  integer. 

dyq         -  i  cfy 
.\  qyq  l-^-  =  pup  l — • 

(XitJb  (JjJb 

dy  _p     up_1  du 
dx      q     yq~l  dx 

_p    up~l    du 
~ of  p\q-xdx 

_p    up~l    du 

q  -p  dx 

Up  '  u  q 

p  ?_i  du 

=  -Uq       — 

q         dx 

=  nu    L  — 
dx 

/.  — —  =  mi"'1—  if  n  is  a  positive  fraction. 

CIX  (XX 

Next,  suppose  that  n  is  negative,  =  —  m,  either  integral  or 
fractional. 

Then  y  =  un  =  u~m  where  m  is  positive, 


u" 


54         DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


.  4/_ 

md  •  1      ^     dwm 
wm 1  •  — ■ 

\JjJu                              LvJL/ 

dx 

u2m 

= 

mum:~x  du     , 

,    b\ 

w2?n     da;       J 

= 

—  mw^1  L  — 
dx 

by  Formula  VI, 


=  nun  1 — ,   since  n  =  —  m. 
dx 

dun  du  •  • 

Therefore  —  =  nun~1 —  for  n  an  integer  or  fraction,  posi- 
dx  dx 

tive  or  negative. 

54.  The  way  in  which  these  formulas  may  be  employed  in 
finding  the  derivative  of  a  function  is  illustrated  in  the  follow- 
ing examples : 

Example  1.     In  the  equation  y  =  x4,  find  -^- 

ax 

Let  u  =  x  and  Formula  VII  applies  directly. 

dv      A    o  dx 

.-.  -2-  =  4  x?  — 

(JjJb  tlJj 

=  4  x3,  since  —  =  1  by  Formula  I. 
dx 

dx 
Example  2.     In  the  equation  ?/  =  4  a?3  —  J  a?  + 1,  find  -^« 

* 

cfa/  _  d  (4  a;3  —  \  x  +  1) 

\AiJb  (J.'Jb 

g d4af  +  d(-js)  +  d^l  by  Formula  ni; 
dx  dx  dx 

=  4  dfl?  _  1  "*5  +  ^1    by  Formula  Vj 

da;       2  da;       da; 
=  12  a;2  -  -,  by  Formulas  VII,  I  and  II. 

Li 


GENERAL  FORMULAS  FOR  DIFFERENTIATION       55 

Example  3.     In  the  equation  y  =  3  Va?  +  - — =  +  4  V^3,  find 
dy  2^Jx 


dy  _  o  dVx     1     yv_g/   ,   .  dVx? 
dx~       dx        2      dx  dx 

1  13 

_o  dx*     1  daQ      .  da^ 
da?       2  cZx  da? 

=  f  x~^  —  \  x~%  +  6  a;*,  by  Formula  VII, 
—  +  6  Va> 


2  Va?     4  Va;3 


Example  4.     In  the  equation  y=(x  —  1)  yV  + 1,  find  ^. 

da; 

ft  =  (0  - 1)  lV^+1  +  V?+I  d("-1),  by  Formula  IV. 
dx  dx  dx 


Leta;2  +  l  =  W.     ...   d  V^  4- 1  =  gvj*  =  1  „-*  gg 

da?  dec        2         da? 

-IftjMyidW-l) 

2  v   T  ;  da; 


vV  + 1 


da;      V^2  +  1 

2  cc2  -  a;  + 1 


Va7+1 

EXERCISES 

Find  -^  in  each  of  the  following  cases : 
dx 

-     „,_2  +  3ca;+4frK2  ,  1     ,  _ 

4x  2X2 

x^x*-x  1.12 


#3  6x*      6««      3^ 


5Q  DIFFERENTIAL   AND  INTEGRAL   CALCULUS 


1  3 


3.    y  = ! ! —  Ans.   -x 1 

4  x*      2x*      x 


x  2" 


4.  y  =  (l+2x+3x2+4;Xs)(l-x)2.  Ans.  -20z3(l-a;) 

5.  y  =  (2  Xs  - 1)  (1  +  ^)2-  -4ns.  18  x5^  + 1) 

„  „,          ea?  .               c(a;  — 2) 

6.  y  =  —      .  •  Ans.   ^ — —^ 


Vx-1  2(x-l)Vx-l 

w                ex                                             A               c  (x  -f-  2) 
7.    y  =  —  •  Ans.   ^ — ! — = 


Vx  +  1  2(a?  +  l)Va;-f-l 


8.   y  =  \ Ans.   

*x  +  l  ^--L^A, 


+  i  0  +  i)Vz2-i 

/aj  +  1 

*#  —  1 


9.    w  =  */^LX±.  ^4ns 


(a?  —  1)V#2  —  1 


V«  +  a?                         ,                      Va  (aj  —  a) 
10.    y  =  — tz — ! — —  ^4ws.   ^ —       v dr 


-y/a  +  -y/x  2  V#  Va  +  x  ( Va  +  Va/)3 

11  V«-V^.  ^s    -Va(Va+Va;) 


Va-ftc  2^Jx(a  +  x)* 


I  1  —  x  A  x  —  2 

12.  y=\~ —  Ans- 


(1  +  ^)3  V(l  -  x)  (1  +  »)' 

13.    y  =  Vl  —  a?  ~\/l  +  x.  Ans. 


14.    y  =  (x2  —  1)  ^/a;3  +  1.  Ans 


6V(l-a;)3(l  +  a;)4 
7  a;4  —  3  x2  +  4  a; 


35 


2  Va3  +  1 
1 


15.  y  =  —  ^4ws 

vr^2"  (i-a^)Vi 

1  —  a?  ^  1  +  x 

16.  y  =    x       — .  Ans. 


x 


Vl  +  a;2  (l  +  ^)Vl  + 

-\/x  A  1  —  3  x2 

17.    y  =  — i — -.  Ans. 


x 


1  +  z2  2Va;(l  +  aj2)5 


GENERAL  FORMULAS  FOR  DIFFERENTIATION      51 


18.  w  = •  Ans.  — _         

a-Vl  +  ar*  Vl  +  a;2(a;-Vl  +  a;2) 

2ar  +  l  ,        -2^  +  11^-9 

19.  gj^JL±^.  Ans.  — 

x(x*  +  3y  3a%2  +  3)* 


20.  y  =  ^(^  +  3)'.  Ans.  V*  +  3(3*-**-3). 

J  a?  - 1  (a?  - 1)2 

In  the  two  following  examples,  find  -^,  — ,  and  then  -f- 
(See  Art.  49.) 

21.  y  =  z3  —  3  z,  z  =  x2  —  x. 

Ans.  ^-=3(z2-l):  —=2x-l:  ^-=3(xi-2x3+x2-l)(2x-l). 
dz       v         ;'  cto  eta       v  /v  y 

22.  y  =  3z2  —  z  +  l,  z  =  x?-l. 

Ans.  ty=6z-l;  —  =  2x;  ^  =  2(6x2-7)x. 
dz  dx  dx 

23 .  In  the  curves  y  =  x3  —  12x2-\-4:X  and  y  —  x3  —  8 x2  —  8, 
given  that  the  abscissas  are  changing  at  the  same  rate,  what 
are  the  coordinates  of  the  points  at  which  the  ordinates  are 
changing  at  the  same  rate  ?  Ans.  (J,  —  J)  ;  (i   —  3-g). 

24.  In  the  curves  of  Exercise  23,  what  is  the  ratio  of  the 
rate  of  change  of  the  ordinate  of  the  first  to  that  of  the  second 
at  the  points  where  the  curves  cut  each  other  ? 

Ans.  If  at  (2,  -  32)  ;  fi  at  (- 1,  -  17). 

25.  At  what  angles  does  the  parabola  #2  =  4a?/  intersect  the 

8  a3 
witch  y  ==  — — -—  ?  Ans.  tan-1  ±  3. 

x2  +  4  a2 

26.  Find  the  points  on  the  curve  y  =  (x  + 1)  (x  —  2)  (x  —  3) 
at  which  the  tangent  line  to  the  curve  is  parallel  to  the  a>axis. 
Makes  45°  with  the  avaxis. 

Ans    fidt^ll     70-26Vl3\      A-VJ3     70  +  26Vl3\ 
3  27         y    \      3       '  27         J' 

(0,  6);    (|,   -|f). 


58  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

27.  If  (x  —  d)  occurs  n  times  as  a  factor  in  a  function  f(x), 

df(x) 
show  that  it  occurs  (n  —  T)  times  in       ,      •     Hence  devise  a 

method  for  determining  whether  a  double  root  occurs  in  an 
algebraic  equation. 

28.  One  end  of  a  rope  is  attached  to  a  boat  at  the  surface 
of  the  water  while  the  other  end  passes  over  a  pulley  10  feet 
above  the  surface.  If  the  rope  is  being  shortened  at  the  rate 
of  40  yards  per  minute,  how  fast  is  the  boat  moving  through 
the  water  when  it  is  5  feet  from  the  foot  of  the  perpendicular 
from  the  pulley  to  the  water  ?  Ans.    2  a/5  ft.  per  sec. 

29.  A  man  is  walking  at  the  rate  of  4  miles  per  hour  on  a 
straight  road  from  which  a  given  point  is  distant  100  feet. 
At  what  rate  is  he  approaching  the  point  when  he  is  200  feet 
distant  from  the  foot  of  the  perpendicular  from  the  point  to 
the  road  ?  Ans.   5.2  ft.  per  sec. 

30.  One  ship  is  41  miles  due  north  of  another.  The  first 
is  sailing  south  at  the  rate  of  8  miles  per  hour,  and  the  second 
is  sailing  east  at  the  rate  of  10  miles  per  hour.  How  rapidly 
are  they  approaching  each  other  ?  How  long  will  they  con- 
tinue to  do  so  ?  Ans.   8  mi.  per  hr.     For  2  hr. 

31.  Two  railroad  tracks  intersect  at  a  station  at  an  angle  of 
60°.  One  train  leaves  the  station  on  one  track  and  travels  at 
the  rate  of  30  miles  per  hour,  and  two  minutes  later  another 
leaves  on  the  other  track  and  travels  at  the  rate  of  60  miles 
per  hour.  At  what  rate  are  the  trains  separating  10  minutes 
after  the  first  started  ?  Ans.    51.4  mi.  per  hr. 

32.  A  reservoir  in  the  form  of  the  frustum  of  an  inverted 
regular  tetrahedron,  whose  lower  edges  are  each  100  feet,  and 
whose  sides  are  inclined  at  an  angle  of  45°  with  the  horizontal, 
is  used  to  supply  a  town  with  water.  If  the  depth  of  the 
water  at  any  instant  is  10  feet,  and  is  falling  at  the  rate  of 
3  feet  per  day,  at  what  rate  is  the  town  being  supplied  ? 

Ans.   23548.8  cu.  ft.  per  day. 


CHAPTER   V 

SUCCESSIVE  DIFFERENTIATION 

55.  The  limit  of  the  ratio  of  the  increment  of  a  function  of 
one  variable  to  the  increment  of  the  variable,  as  the  increment 
of  the  variable  approaches  zero,  is  the  derivative  of  the  func- 
tion with  respect  to  the  variable.  This  derivative  is  called 
the  first  derivative  of  the  function  with  respect  to  the  variable. 
Now  this  derivative  is  usually  itself  a  function  of  the  variable 
and  can  be  differentiated  with  respect  to  it.  The  derivative 
of  the  first  derivative  is  called  the  second  derivative  of  the 
function  with  respect  to  the  variable.  The  second  derivative 
is  usually  also  a  function  of  the  variable  and  can  be  differen- 
tiated with  respect  to  it.  The  derivative  of  the  second  deriva- 
tive is  called  the  third  derivative  of  the  function  with  respect 
to  the  variable.  And  so  on.  In  general,  the  derivative  with 
respect  to  the  variable  of  the  (n  —  l)th  derivative  of  a  function 
with  respect  to  that  variable  is  called  the  nth  derivative  of  the 
function  with  respect  to  that  variable. 

For  example,  in  the  equation  y  =  x*, 

dv 

~  =  4:X3,  the  first  derivative  of  y  with  respect  to  x ; 

d  fdy\ 
— (  -^  J  =  12  x2,  the  second  derivative  of  y  with  respect  to  x  ; 

-^-(  -rH  ^  )  )  =  24  sc,  the  third  derivative  of  y  with  respect  to  x: 
dx\dx\dxJJ  '  v  i  > 

— [  — [  — (  —  ] ) )  =  24,  the  fourth  derivative  of  y  with  respect  to  x. 
dx\dx\dx\dxj  J  J 

Since  24  is  a  constant,  the  fifth  and  all  succeeding  deriva- 
tives are  zero. 

59 


60  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

56.    Notation.     In  the  equation  y  =f(x)} 

-^       is  denoted  sometimes  by  f'(x)  > 

(IX 

jL(cly)     is  denoted  by  ^  or  sometimes  by  f"(x)  ; 
dx  \dxj  dx2 

— /  — (JL  ]  ]  is  denoted  by  — \  or  sometimes  by  f'"(x). 
dx  \dx  \dxj  J  dx3 

dnv 
The  nth  derivative  of  y  with  respect  to  x  is  denoted  by  —J 

or  sometimes  by/(w)(#). 

EXERCISES 

1.  In  the  equation  y  =  x?,  find  — ^-  Ans.  60  x2. 

2.  In  the  equation  y  =  -,  find  J-  J.?is.  — r- 

&  dec2  a;3 

3.  In  the  equation  y  =  2xs  —  3x2  —  12x  +  l,  find  the  values 
of  x  that  make  f'(x)  zero  and /"(a?)  positive.  How  many  such 
values  are  there  ?  Ans.  2.     One  value. 

4.  In  the  equation  y  =  xs  —  6  a^  +  a—  1,  find  /'(2),  f"(l), 
/"(- 1),  /'(-  2).  Ans.  - 11,  -  6,  - 18,  37. 

5.  In  the  equation  y  =  10  x6  +  36  x5  -  75  x4  -  300  x?  + 120  x2 
+  720  a  + 1,  find  the  value  or  values  of  x,  (1)  that  make  f'(x) 
zero  and  f"(x)  positive;  (2)  that  make  /'(&)  zero  and  f"(x) 
negative;  (3)  that  make  flY(x)  zero  and  fY(x)  positive; 
(4)  that  make  /^(x)  zero  and  fY(x)  negative. 

Ans,    (i)  _3,  -1,  2;  (2)   -2,  1;_ 

,q.   -6  +  V86.    m   -6-V86 


CHAPTER  VI 

APPLICATIONS  OP  DERIVATIVES  TO  CURVES. 
MAXIMA  AND  MINIMA 

57.  In  the  curve  whose  equation  is  y  =f(x),  x  is  the  abscissa, 
and  y  or  f(x)  the  ordinate  of  any  point  (x,  y)  on  the  curve. 
Suppose  that  f(x)  is  a  single  valued  and  continuous  function 
of  x.  If  x  continually  increases,  the  point  (x,  y)  moves  on  the 
curve  so  that  its  projection  on  the  x-axis  moves  from  left  to 
right.  If  f{x)  continually  increases  as  x  increases,  the  ordi- 
nate of  the  curve  corresponding  to  the  abscissa  x  is  continually 
increasing  as  the  curve  is  being  drawn  from  left  to  right. 
That  is,  the  curve  is  rising  when  drawn  from  left  to  right. 
If  f(x)  continually  decreases  as  x  increases,  the  ordinate  of 
the  curve  corresponding  to  the  abscissa  x  is  continually  de- 
creasing as  the  curve  is  being  drawn  from  left  to  right.  That 
is,  the  curve  is  falling  when  drawn  from  left  to  right.  If  f(x) 
is  neither  increasing  nor  decreasing  as  x  increases,  the  curve  is 
neither  rising  nor  falling  and  must  therefore  be  parallel  to  the 
sc-axis.  In  this  case  the  equation  becomes  y  =  c,  where  c  is 
constant,  which  is  the  equation  of  a  line  parallel  to  the  a?-axis. 

We  can  thus  think  of  the  function  f(x)  as  increasing, 
decreasing,  or  remaining  constant,  as  x  increases,  if  the  thought 
is  merely  of  the  function  as  given  by  the  equation  y  =f(x), 
or  of  the  curve  as  rising,  falling,  or  remaining  parallel  to  the 
a>axis~  when  drawn  from  left  to  right,  if  the  thought  is  of 
the  curve  that  represents  the  equation  y=f(x). 

58.  Let  us  consider  the  curve  whose  equation  is  y  =  x3  —  4  x2 
-f  4:X  —  1  and  determine  the  values  of  x  between  which  the 

61 


62  DIFFERENTIAL  AND  INTEGRAL    CALCULUS 

curve  is  rising  or  falling,  or  where  it  is  turning  from  rising  to 
falling  or  from  falling  to  rising,  when  drawn  from  left  to  right. 
Let  x  =  x0.  To  determine  whether  the  curve  is  rising  or 
falling  as  x  increases  through  x0,  we  shall  calculate  the  ordi- 
nates  corresponding  to  x0  and  xQ  +  Ax,  and  compare  the  two 
in  length. 

Let  x  =  xQ.     .-.  y0  —  x0s  —  4  x02  +  4  xQ  —  1. 

Let  x  =  x0  +  Aa;. 

.-.  y0  +  Ay  =  (x0  -h  Ax)3  -  4  (x0  +  Ax)2  +  4  (a;0  +  Aa;)  - 1. 

.-.  Ay  =  (3  a;02  —  8  xQ  +  4)  A  x  +  (3  a;0  —  4)  Ax  +  Aa?. 

.-.  ^  =  (3a;02-8a;0-J-4)  +  (3a;0-4)Aa;  +  Aa?. 
Aa; 

By  taking  Aa;  small  enough,  — ^  can  be  made  to  have  a  value 

as  near  3  a?02  —  8  x0  +  4  as  we  please.     Therefore,  if  Aa,*  is  small 

enough,  — ^  will  have  the  same  sign  as  3  x02  —  8  x0  +  4.     Sup- 
Ax 

pose  that  3  #02  —  8  xQ  +  4  is  positive.     Therefore,  if  Aa;  is  small 

enough,  — ^  is  positive.     Now  Aa;  is  positive  since  by  supposi- 

Aa; 

tion  the  variable  is  increasing.  Therefore  Ay  is  positive.  There- 
fore y0+ Ay  is  greater  than  y0.  Therefore,  as  Aa;  increases  from 
zero,  or  as  x  increases  from  x0,  the  ordinate  of  the  curve  is 
increasing.  That  is,  the  curve  is  rising.  Therefore  if  x0,  is 
such  that  3  a;02  —  8  x0  +  4  is  positive,  the  curve  is  rising  at 
x  =  x0,  when  drawn  from  left  to  right. 

The  student  can  readily  make  the  necessary  changes  in  the 
proof  and  show  that  if  3  x02  —  8  x0  +  4  is  negative,  the  curve  is 
falling  at  x  =  x0%  when  drawn  from  left  to  right. 

The  values  of  x  between  which  3  a;02  —  8  x0  -f  4  is  positive  or 
negative  can  readily  be  found  by  factoring  the  expression 
3  x02  —  8  x0  -f-  4.     The  factors  are  3  x0  —  2  and  x0  —  2. 

If  x0  is  less  than  -|,  both  factors  are  negative,  and  therefore 
their  product  is  positive.     If  x0  is  greater  than  J  and  less  than 


APPLICATIONS   OF  DERIVATIVES   TO   CURVES       63 


2,  3  x0  —  2  is  positive  and  xQ  —  2  is  negative,  and  therefore  their 

product  is  negative.     If  x0  is  greater  than  2,  both  factors  are 

positive,  and  therefore  their  product  is  positive.     The  curve  is 

therefore : 

rising  for  all  values  of  x  less  than  J ; 

falling  for  all  values  of  x  between  -|  and  2 ; 

rising  for  all  values  of  x  greater  than  2. 

The  curve,  when  drawn  from  left  to  right,  is  turning  at 
x  —  f  from  rising  to  falling,  and  at  x  =  2  from  falling  to  rising. 
At  these  values  3  x02  —  8  x0  +  4  is  zero. 

59.   In  the  equation  we  have  been  considering,  namely, 

y  =  XS—    4:  X2  -\-  4:  X  —  1, 

when  x  =  xQ. 


(3  x<?  -  8  x0  +  4)  +  (3  x0-  4) Ax  +  Ax2  =  ^ 

i^X 


...3^-8*0  +  4  =  ^0 


dx 


'Ay 


when  a?  =  as 


o> 


We  therefore  see  that  in  this  particular  problem  as  a;  increases 


through  the  value  x0)  if 


#  =  xa,  and  if 


da; 


dy 
dx 


is  positive  the  curve  is  rising  at 


is  negative,  the  curve  is  falling  at  x  =  x0. 


60.   In  general,  in  any  equation  y=f(x)   where  f{x)  is  a 
single  valued  and  continuous  function  of   x,  as  x  increases 

is  positive,  the  curve  that  represents  the 


through  x0,  if  -^ 
dx 


dy 


is  negative,  the  curve  is  falling 


equation  is  rising,  and  if  ~^- 
,  dx 

when  x  =  x0. 

This  theorem  can  be  readily  established  as  follows: 

From    the  definition  of   a  limit   the  difference   between  a 

function  and  its  limit  is  infinitesimal.     Therefore,  since  -^ 

dv  Ax 

when  x  =  x0  is  a  variable,  and  -^ 

dx 


Ay 

Ax 


dy 
dx 


its  limit, 
=  e,  where  e  is  infinitesimal. 


64 


DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


Suppose  that  -^ 

LttJO 


is  not  zero. 


By  taking  Ax  small  enough,  e  can  be  made  to  have  a  value 
as  near  zero  as  we  please.     Therefore,  if  Ax  is  small  enough, 

— ^  when  x  =  x0  will  have  the  same  sign  as  — 

is  positive. 


Suppose  that  — 

(XX    «_„ 


Ay 


Suppose  that  — 
^  dx 

If  Ax  is  small  enough,  -^  when 

Ax 


If  Ax  is  small  enough,  —  when  x  =  x0  is  positive.  Now  Ax 
is  positive  since  by  supposition  the  variable  is  increasing. 
Therefore  Ay  is  positive.  Since  Ay  is  positive,  y0  +  Ay  is  greater 
than  y0.  Therefore  as  Ax  increases  from  zero,  or  as  x  increases 
from  xQ,  the  ordinate  of  the  curve  is  increasing.  Therefore  at 
x  =  x0  the  curve  when  drawn  from  left  to  right  is  rising. 

is  negative. 

x  =  x0  is  negative.     Now  Ax 

is  positive  since  by  supposition  the  variable  is  increasing. 
Therefore  Ay  is  negative.  Since  Ay  is  negative,  y0  4-  Ay  is  less 
than  y0.  Therefore  as  Ax  increases  from  zero  or  as  x  increases 
from  x0,  the  ordinate  of  the  curve  is  decreasing.  Therefore  at 
x  =  xQ  the  curve  when  drawn  from  left  to  right  is  falling. 

61.  When  the  curve  is  drawn  from  left  to  right,  the  slope  of 
the  tangent  line  may  be  increasing  or  decreasing  while  either 
positive  or  negative. 

If  it  is  increasing  and  positive,  the  tangent  line  will  take  a 
succession  of  positions  in  the  order  1,  2,  3,  4  of  Fig.  15,  and 


12      3         4 
Fig.  16. 


APPLICATIONS   OF  DERIVATIVES   TO   CURVES       65 

the  curve  will  be  as  drawn  in  Fig.  15.  If  it  is  increasing  and 
negative,  the  tangent  line  will  take  a  succession  of  positions  in 
the  order  1,  2,  3,  4  of  Fig.  16,  and  the  curve  will  be  as  drawn 
in  Fig.  16.  In  either  case  the  curve  is  said  to  be  concave 
upwards. 

If  the  slope  of  the  tangent  line  is  decreasing  and  positive, 
the  tangent  line  will  take  a  succession  of  positions  in  the  order 
1,  2,  3,  4  of  Fig.  17,  and  the  curve  will  be  drawn  as  in  Fig.  17. 


If  it  is  decreasing  and  negative,  the  tangent  line  will  take 
a  succession  of  positions  in  the  order  1,  2,  3,  4  of  Fig.  18,  and 
the  curve  will  be  as  drawn  as  in  Fig.  18.  In  either  case  the 
curve  is  said  to  be  concave  downwards. 

62.  In  Art.  58,  we  found  a  method  for  determining  the 
values  of  x  between  which  the  curve  y  =  xs  —  4  a2  +  4  #  —  1  is 
rising  or  falling  or  for  which  it  is  neither  rising  nor  falling, 
when  drawn  from  left  to  right.  We  shall  now  find  a  method 
for  determining  the  values  of  x  in  the  same  curve  between 
which  the  curve  is  concave  npwards  or  concave  downwards  or 
for  which  it  is  neither  concave  upwards  nor  concave  downwards. 

If  y=x*-4;x2+4;x-l, 

then  ^  =  3o2-8a>  +  4, 

dx 

and  J  =  6  x  —  8, 

dxr 


66  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

The  expression  3  x2  —  8  #  +  4  is  a  function  of  x.     Call  it  u. 

Then  if  —  is  positive,  the  curve  whoSe  equation  is  u  =  3  x2 
dx 

—  8  #  4-  4  is  rising ;  that  is,  u  is  increasing.    (See  Art.  60.)    Now 

r/?/        d  n  d  n  ■  • 

—  is  — j|-     Therefore  if  — -^  is  positive,  u  is  increasing.    But  u, 
fjjju        iXQcr  (X/iX/ 

or  -^,  is  the  slope  of  the  tangent  line  to  the  curve  at  a  point 

(x,  y)  on  the  curve.    (See  Art.  39.)    Therefore,  if  — ^  is  positive, 

the  slope  of  the  tangent  line  to  the  curve  at  a  point  on  the 
curve  is  increasing.     Therefore  the  curve  is  concave  upwards. 

A  similar  proof  will  show  that  if  — %  is  negative,  the  curve 

dx" 
is  concave  downwards. 

The  expression  6  x  —  8  is  negative  if  x  is  less  than  |-  and 

positive  if  x  is  greater  than  f .     The  curve  is  therefore  concave 

downwards  for  all  values  of  x  less  than  |-  and  concave  upwards 

for  all  values  of  x  greater  than  J. 

63.   We  saw  in  Art.  58  that,  in  the  curve  y=xz— 4a^4-4cc— 1, 
for  values  of  x : 

less  than  -|,  the  curve  is  rising ; 
between  -|  and  2,  the  curve  is  falling ; 
greater  than  2,  the  curve  is  rising ; 

and  in  the  preceding  article  that,  for  values  of  x : 

less  than  ^,  the  curve  is  concave  downwards ; 
greater  than  |-,  the  curve  is  concave  upwards. 

From  these  results,  we  can  plot  the  curve  by  plotting  a  few 

points  and  drawing  the  curve  through  them  to  satisfy  the 

above  data. 

The  ordinates  whose  abscissas  are  f,  f ,  and  2  are  -fj,  —  H?  ano^ 

—  1  respectively.     Plot  these  points.     They  are  D,  F,  and  G 

3  —  V5 
in  the  figure.    When  x  =  0,  y  =  —  l.     When  y  =  0,x=  — - — , 

±f  and  ?  +  V^.     Plot  these  points.     They  are  B,  0,  E,  and  H 


APPLICATIONS   OF  DERIVATIVES   TO   CURVES       07 


in  the  figure.  When  x  =  —  1,  y  =  —  10.  When  x  =  3,  y  =  2. 
Plot  these  points.  They 
are  A  and  K  in  the  figure. 
Draw  a  curve  concave  down- 
wards through  ABCDEF, 
then  concave  upwards 
through  GHK.  We  thus 
have  the  curve  drawn  be- 
tween x  =  —  1  and  x  =  3. 

64.  In   Art.    62,  we   saw 
that,  in  the  curve  y=x3— 4  x2 

-\-4:X  —  1,  if  — ^  is  positive, 
dxr 

the  curve  is  concave  up- 
wards, and  if  negative,  concave  downwards.  That  this  is  true 
for  any  curve  y=f(x)f  where  f(x)  is  a  single  valued  and  con- 
tinuous function  of  x,  can  readily  be  shown. 

In  the  equation  y  =f(x),  -^  is  a  function  of  x.     Then  if  its 

.-72  ,7 

derivative,  namely  —  -9,  is  positive,  —  is  increasing  and  the 

curve  is  concave  upwards.    If  — ^  is  negative,  —  is  decreasing 

ctx  (XX 

and  the  curve  is  concave  downwards. 

65.  Points  of  inflexion.  The  points  on  the  curve  at  which 
the  curve  turns  from  concavity  upwards  to  concavity  downwards 
or  vice  versa  are  called  points  of  inflexion  of  the  curve. 

Thus,  in  the  curve  y  =  x3  —  4  x2  +  4  x  —  1,  the  point  (-J,  —  \\) 
is  a  point  of  inflexion. 


when  -^ 


66.    In  Art.  58,  we  saw  that  in  the  curve  y  =  x3  —  4#2-{-4a;— 1, 
or  3  a?02  —  8  a?0  +  4  is  zero,  the  curve  is  turning  from 

rising  to  falling  or  from  falling  to  rising.    It  is  not  true,  however, 
that  in  any  curve  y=f(x)  when  — "  i       is  zero  the  curve  is  turning 


68 


DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


from  rising  to  falling  or  from  falling  to  rising.     It  may  be,  as 
in  Fig.  20  or  Fig.  21,  where  it  is  rising  or  falling,  and  at  a  par- 


0 


Fig.  20. 


Fig.  21. 


ticular  point  x  =  x0  has  a  value  at  which  the  tangent  line  is 
parallel  to  the  a>axis. 

As  an  illustration,  consider  the  curve 

-  x*  -  6  x2  +  8  x  + 1. 


y 

^=±x> 

dx 


12  a2 +  8. 


g  =  12(rf-l). 

For  values  of  x  in  the  neighborhood  of  1,  -^  is  positive,  and 

dx 

therefore  for  these  values  the  curve  is  rising.     When  x  =  1, 
-^  is  zero,  and  therefore  the  tangent  line  to  the  curve  at  this 

point   is    parallel    to    the   #-axis. 

For  values  of  x  immediately  less 

d  ?/ 
than  1,  —^-  is  negative,  and  there- 


Y 


dx2 
fore  for  these  values  the  curve  is 

concave  downwards.      For  values 
of  x  immediately  greater  than  1, 

Fig.  22.  —  is  positive,  and  therefore  for 

dx2 

these  values  the  curve  is  concave  upwards.     In  the  neighbor- 
hood of  x  =  1  it  is  therefore  as  in  Fig.  22. 

67.    Definitions.     If  a  curve  rises  to  a  certain  point  and  then 
falls,  that  point  is  called  a  maximum  point  of  the  curve. 


APPLICATIONS   OF  DERIVATIVES   TO   CURVES       69 


If  a  curve  falls  to  a  certain  point  and  then  rises,  that  point 
is  called  a  minimum  point  of  the  curve. 

Thus,  in  the  curve  y  =  x3  —  4  x2  +  4  x  —  1,  (f,  -fj)  is  a  maxi- 
mum point  and  (2,  —  1)  is  a  minimum  point  of  the  curve. 

68.   We  saw  that,  if  —  is  positive,  the  curve  is  rising,  and 

dx 

if  negative,  falling  when  drawn  from  left  to  right.     Then,  at  a 

maximum  point,  -^  changes  from  positive  to  negative.     Now 
dx 

a  continuous  function  can  change  in  sign  only  by  passing 
through  the  value  zero,  and  a  function  which  has  only  infinite 
discontinuities,  by  passing  through  the  value  zero  or  by  becom- 
ing infinite.     Therefore,  at  a  maximum  point,  -^  is  zero  or 


Fig.  23 


Fig.  24. 


infinite.  The  curve  at  a  maximum  point  is  therefore  shaped 
as  in  Fig.  23,  where  the  tangent  line  is  parallel  to  the  a>axis, 
or  as  in  Fig.  24,  where  it  is  perpendicular  to  the  a>axis. 

The  maximum  point  at  which  -^  is  zero  may  be  called  an 

dx 

ordinary  maximum  point.     It  is  the  kind  that  appears  in  the 


curve  of  Art.  63 

is  called  a  cusp  maximum  point. 

Similarly  we  may  have  an  ordi- 
nary minimum  point  of  the  kind 
that  appears  in  Art.  63,  and  a  cusp 

minimum  point  at  which  -^  is  in- 

dx 

finite.     In  the  latter  case  the  curve 
appears  at  the  point,  as  in  Fig.  25. 


The  maximum  point  at  which  -^  is  infinite 

dx 


0 


Fig.  25. 


70         DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


Tests  for  Maximum  or  Minimum  Values 

69.   We  saw  that,  if  -^  is  zero  for  x  =  x0,  and  if  -^  is  posi- 

dx  ax 

tive  before  x  =  x0  and  negative  after  x  =  x0,  the  point  (x0,  y0) 
on  the  curve  is  a  maximum  point. 

In  this  condition  for  a  maximum  point,  the  conditions  that 


-^  is  positive  before  x  =  x0  and  negative  after  x  =  x0  may  be 
expressed  in  the  one  condition,  — J         is  negative. 


For,  suppose  that  —4  is  negative  when  a;  =  #0-     Since   -4, 
die2  d 

the  derivative  with  respect  to  x  of  -^,  is  negative  when  £  =  ic0, 

7  c^  d.v 

-^  is  decreasing  as  as  passes  through  the  value  x0,  and  since  -^ 
eZa?  dx 

is  zero  when  #  =  #<>,  it  must  have  been  positive  before  x  =  x0 
and  negative  after  x  =  x0. 

We  can  therefore  conclude  that,  if 


=  0, 


dy 
dx 

d  v 
and  -\ 

dx1 

the  point  (x0,  y0)  on  the  curve  is  a  maximum  point. 
Similarly,  if  , 

dx 


o, 


d?v 

and  d4 


=  +, 


the  point  (x0,  y0)  on  the  curve  is  a  minimum  point. 

70.  The  above  reasoning  is  based  on  the  supposition  that 
the  function  f(x)  is  single-valued  and  continuous  for  all  the 
values  of  x  considered. 

If  the  function  f(x)  is  not  single-valued  it  can  be  made  so, 
as  explained  in  Art.  15. 


MAXIMA  AND  MINIMA  71 

If  the  function  is  infinite  or  otherwise  discontinuous  for 
certain  values  of  x,  these  values  may  be  avoided  in  the  dis- 
cussion. For  all  other  values  the  reasoning  of  the  preceding 
articles  will  hold. 

71.  As  an  illustration  of  the  method  of  plotting  a  curve 
when  the  function  which  it  represents  is  infinite  for  certain 
values  of  the  variable,  consider  the  curve 

x(x  —  l) 
U —     • 

jj  X  —  o 

For  x  =  f,  the  function  is  infinite  or  infinite  negatively,  so 

that  this  value  of  x  will  be  avoided  in  the  discussion.     For  all 

other  values  of  x, 

dy  _2x2  —  6x-\-3 

dx~     (2x-3f    ' 


and 


d2y  _         6 


dx2     (2x-  3)J 


Factor  the  numerator  of  — — 6x  +  S. 

(2x-  3)2 

•    *V-     t  g LI  2       J 

"  dx  (2x-3f 

An  examination  of  this  expression  shows  that  -^  is : 

_  dx 

positive  for  all  values  of  x  less  than     ~        • 

negative  for  all  values  of  x  between     ~^     and       ' 

.3  2  2' 

avoiding  - ; 

Li  

positive  for  all  values  of  x  greater  than      *"        ;  and 

zero  when  x  =    ~        and  x  =    ~*~       . 

2  2 


72         DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

The  curve  is  therefore  : 

rising  for  all  values  of  x  less  than  — - — ; 

falling  for  all  values  of  x  between  — ^ —  and 


avoiding  - ; 
rising  for  all  values  of  x  greater  than 


2 
3+V3 


An  examination  of 


^-^  shows  that  g  is: 


negative  for  all  values  of  x  less  than  - ; 

3 

positive  for  all  values  of  x  greater  than  -. 

The  curve  is  therefore  : 

3 
concave  downwards  for  all  values  of  x  less  than  - ; 


concave  upwards  for  all  values  of  x  greater  than  - ; 

3  —  V3 

and  has  a  maximum  point  when  x  =  — - — ,  and  a  minimum 


point  when  x  = 


3  +  V3 


An  examination  of  -g ~  shows  that  the  curve  cuts  the 

aj-axis  when  x  =  0  and  x  =  1, 
and  that 


Y 


limit 

X=—  CO 

^x  (x  —  1) 
_2a-3 

=  —  oo, 

and 

limit 

£  =  00 

x  (x  —  1) 
_2a-3  _ 

=  +  Q0. 

The  curve  is  therefore  as 

drawn  in 

Fig.  26. 

Fig.  26. 

Since  -^  cannot  become  zero  for  any  value  of  x,  this  curve 
cix 

has  no  point  of  inflection. 


MAXIMA  AND  MINIMA 


73 


EXERCISES 

1.    Draw  a  branch  of  a  curve  at  every  point  of  which : 

dy  .  .,.  ,  d2y  . 

(a)  x  is  positive,  y  is  positive,  -f-  is  positive,  and  -=-g  ls  positive; 

dv  d"v 

(b)  x  is  positive,  y  is  negative,  -=-  is  positive,  and  -j-^  is  positive ; 

dy  d2y 

(c)  x  is  negative,  y  is  positive,  ~  is  negative,  and  -v4  is  negative ; 

(d)  x  is  negative,  y  is  positive,  ~  is  negative,  and  -^§  is  positive. 


d2v 
2.    In  the   curve   y=f(x),  show  that   if   -^ 


d3y 
dx3 


dx- 


=  0,  and 


+,  the  curve  has  a  point  of  inflection  when  x  =  x0. 

=  0  also,  how  is  the  curve  situ- 


3.    In  Exercise  2,  if 


dx 

ated  at  the  point  of  inflection  ? 

Arts.   The  curve  is  parallel  to  the  #-axis. 


dv 
4.   In  the  curve  y  =f(x),  if  -j- 


and 


dhj 


=  0, 


dx2 


=  0, 


d3y 
dx* 


o, 


d#4 
when  &  =  % 


=  +,  show  that  the  curve  has  a  minimum  point 


Investigate  the  following  curves  for  ranges  of  values  of  x  in 
which  the  curves  are :  (1)  rising ;  (2)  falling ;  (3)  concave 
upwards ;  (4)  concave  downwards ;  and  for  values  of  x  for 
which  they  have  :  (5)  maximum  points ;  (6)  minimum  points ; 
(7)  points  of  inflection.     In  each  case,  plot  the  curve. 

5.  y  =  x?  —  2x  +  4.  10.  y  =  x*. 

6.  y  =  x(x  —  l)(x  —  2).  11.  y  =  x\ 

7.  y  =  (x—l)(x—2)(x—3).  12.  y  =  x(x  —  3)3. 

8.  y  =  x(x2-3).  13.  y  =  x(x2  +  l). 

9.  y  =  x2(x2  —  S).  14.  y  =  x*(x  —  3). 


74         DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

15.  y=*+\ 

16.  y  = 


17.   y  = 


x  +  1 

X—  1 

x  +  2  x  + 


18.    y=     X     ■ 
9      x  +  1 

x(x  +  l) 
19-   *~-- ^=2    • 

20.     ?/  —  v             7  v 

-2) 

Investigate  the  following  curves  for  maximum  and  minimum 
points.     In  each  case  plot  the  curve. 

21.  y2  =  4:x(l  —  x).  26.    y5  =  x  +  2. 

22.  y2  =  x(x2  —  1).  _„       1,4        4 
^          v           J                          27.    cc3  +  2/3  =  a3. 


28.    y  =  (x-l)i  +  4:. 


23.  2/2  =  x3(l  —  x). 

24.  y2  =  xz. 

„      1  29.    y=(x  — 1)*  +  3. 

25.  2/2  =  --  V  y 

72.  Definitions.  If  a  single  valued  and  continuous  function 
increases  to  a  certain  value  and  then  decreases,  that  value  is 
called  a  maximum  value  of  the  function. 

If  a  single  valued  and  continuous  function  decreases  to  a 
certain  value  and  then  increases,  that  value  is  called  a  mini- 
mum value  of  the  function. 

73.  We  saw  in  Art.  57  that  if  the  curve  y=f(x)  is  rising, 
the  function  f(x)  is  increasing,  and  if  falling,  decreasing.  The 
tests  for  a  maximum  or  minimum  value  of  the  function  will 
therefore  be  the  same  as  for  a  maximum  or  minimum  point  of 
the  curve,  namely : 

=  0, 


If 
and 


dy 
dx 


cVy 
dx2 


as  x  increases  through  the  value  x0,  the  function  f(x)  has  a 
maximum  value  when  x  =  x0 ;  and 


MAXIMA   AND  MINIMA 


15 


If 


and 


dy 
dx 

d2y 
dx2 


=  0, 


=  +: 


as  x  increases  through  the  value  x0,  the  function  f(x)  has  a 
minimum  value  when  x  =  x0. 

74.  As  an  illustration  of  some  of  the  uses  to  which  the  above 
tests  for  maximum  or  minimum  values  of  a  function  may  be 
put,  consider  the  following  example  : 

Example.  Given  that  the  strength  of  a  beam  of  rectangular 
cross  section  is  proportional  to  the  product  of  its  breadth 
by  the  square  of  its  depth,  find  the  breadth  and  depth 
of  the  strongest  rectangular  beam  that  can  be  cut  from  a 
cylindrical  log  which  is  2  a  inches  in 
diameter. 

Let  s,  x,  and  y  denote  the  strength, 
breadth,  and  depth  respectively  of  the 
beam. 

Then  by  supposition,  socxy2.  .*.  s  =  kxy2 
where  k  is  a  constant  depending  on  the 
material  of  which  the  beam  is  composed. 


Fig.  27. 


From  the  geometry  of  the  figure,  x2 -\- y2  =  4:  a2. 
.-.  s  =  kx(4,a2 —  x2). 

.-.  ^-  =  k\4a2~3x2\. 
dx 


d2s 
dx2 


=  —  6  kx. 


When  x= 


2a 


V3 


^  =  0  and  ^  =  -.    Therefore,  when  x=^, 
ax  dx~  ^/g 

Substitute  x—  — -  in 
V3 
Therefore  the   dimen- 

2a 


the  strength  of  the  beam  is  a  maximum. 

ar2  +  2/2  =  4  a2.      Therefore  ?/  =  2aV|. 

sions  of  the  strongest  beam  are:   breadth  ^Zi  depth  2a V¥. 

V3 


76  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

EXERCISES 

1.  G-iven  that  the  stiffness  of  a  beam  of  rectangular  cross 
section  is  proportional  to  the  product  of  its  breadth  and  the 
cube  of  its  depth,  find  the  breadth  and  depth  of  the  beam 
of  greatest  stiffness  that  can  be  cut  from  a  cylindrical  log 
2  a  inches  in  diameter. 

Ans.  Breadth  =  a  in.     Depth  =  a  V3  in. 

2.  A  rectangular  box  with  open  top  and  square  base  is  to 
be  constructed  to  hold  2  gallons.  Find  the  dimensions  of  the 
box  in  inches  which  will  contain  the  least  possible  material. 
(231  cu.  in.  =  1  gal.) 

Ans.  Side  of  base  =  9.74  in.     Height  =  4.87  in. 

3.  A  rectangular  box  with  closed  top  and  square  base  is  to 

be  constructed  to  hold  2  bushels.     The  cost  of  the  material  in 

the  bottom  is  3  cents  per  square  inch,  in  the  ends  2  cents  per 

square  inch,  and  in  the  sides  and  top  1  cent  per  square  inch. 

Find  the  dimensions  of  the  box  of  least  cost.      (2150.42  cu. 

in.  =  1  bu.) 

Ans.  Side  of  base  =  14.77  in.     Height  =  19.71  in. 

4.  A  wall  6  feet  high  is  situated  8  feet  from  the  side  of  a 
house.  What  is  the  length  of  the  shortest  ladder  that  will 
reach  from  the  ground  beyond  the  wall  to  the  house  ? 

Ans.  19.73  ft. 

5.  Show  that  the  maximum  rectangle  inscriptable  in  a 
circle  is  a  square. 

6.  Find  the  right  circular  cone  of  maximum  volume  that 
can  be  inscribed  in  a  sphere  of  radius  R. 

Ans.  Kadius  of  base  =  f  V2  R.     Height  =  §  R. 

7.  Find  the  altitude  of  the  right  circular  cylinder  of  maxi- 
mum surface  that  can  be  inscribed  in  a  sphere  whose  radius  is  R. 

Ans.  [2--^~]2R. 

VS. 


MAXIMA   AND  MINIMA  77 

8.  Find  the  altitude  of  the  right  circular  cylinder  of 
greatest  volume  that  can  be  inscribed  in  a  sphere  whose 
radius  is  R.  ^ns   2  R 

'  vs" 

9.  Find  the  altitude  and  volume  of  the  right  circular  cone 
of  least  volume  that  can  be  circumscribed  about  a  sphere 
whose  radius  is  R.         Ans.  Altitude  =  4  R.     Volume  =  §  ttR?\ 

10.  A  square  piece  of  tin  each  of  whose  sides  is  a  has  a 
small  square  cut  out  of  each  corner.  Find  the  side  of  the 
small  square  that  the  remainder  may  form  a  box  of  maximum 

contents.  ,,        a 

Ans.  -• 

6 

11.  The  work  of  propelling  a  steamer  through  the  water 
varies  as  the  cube  of  her  speed.  Find  the  most  economical 
speed  against  a  current  running  4  miles  per  hour. 

Ans.  6  mi.  per  hr. 

12.  The  cost  of  fuel  consumed  in  propelling  a  steamer 
through  the  water  varies  as  the  cube  of  her  speed,  and  is 
$  25  per  hour  when  the  speed  is  10  miles  per  hour.  The 
other  expenses  are  $  100  per  hour.  Find  the  most  economical 
speed.  Ans.  12.6  mi.  per  hr. 

13.  A  Norman  window  consists  of  a  rectangle  surmounted 
by  a  semicircle.  For  a  given  perimeter,  determine  the  breadth 
and  height  of  the  window  that  gives  a  maximum  quantity  of 

light. 

2  v 
Ans.  Breadth  =  height  of  rectangle  =  — ±— ,  where  p  is  the 

perimeter.  ^  "~ 

14.  The  area  cut  off  from  a  parabola  by  any  double  ordi- 
nate is  |  the  circumscribing  rectangle.  Find  the  maximum 
parabola  which  can  be  cut  off  from  a  right  cone,  in  terms 
of  the  slant  height  of  the  cone  and  the  radius  of  the  base. 

Ans.  ^haV3,  where  a  =  radius  of  base  and  h  =  slant  height. 


78         DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

15.  A  barnyard  gate  is  400  feet  from  a  straight  river. 
The  gate  of  the  pasture  is  100  feet  from  the  river  and  500  feet 
from  the  barnyard  gate.  If  the  cattle  go  to  drink  on  their 
way  to  the  pasture,  determine  the  length  of  their  shortest 
possible  path  from  gate  to  gate.  Ans.  640.31  ft. 

16.  A  miner  wishes  to  dig  a  tunnel  from  a  point  A  to  a 
point  B,  300  feet  below  A  and  500  feet  distant  from  it.  Below 
A  is  bed  rock  and  above  A  is  earth,  the  separating  surface 
being  a  horizontal  plane.  Given  that  the  cost  of  tunneling 
through  earth  is  $  1  and  through  rock  $  3  per  linear  foot,  find 
the  cost  of  a  tunnel  of  minimum  cost.  Ans.  $1248.53. 

17.  A  man's  house  is  situated  20  yards  from  a  straight  car 
track.  The  man  can  walk  at  the  rate  of  4  miles  per  hour  and 
the  car  travels  at  the  rate  of  8  miles  per  hour.  Where  must 
the  man  leave  the  car  in  order  that  he  may  reach  home  in  the 
least  possible  time. 

Ans.  34.64  ft.  from  foot  of  perpendicular  from  house  to  track. 

E 

18.  Assuming  that  the  current  in  a  voltaic  cell  is  C  = — , 

6  r  +  R 

E  being  the  electromotive  force,  r  the  internal  resistance, 
R  the  external  resistance ;  and  that  the  power  given  out  is 
P  =  RG2;  prove  that  P  is  a  maximum  when  r  =  R.  Trace  the 
curve  that  shows  the  variation  of  P  as  R  varies. 

(Perry's  Calculus  for  Engineers.) 

19.  A  length  I  of  wire  is  to  be  cut  into  two  portions,  which 
are  to  be  bent  into  the  forms  of  a  circle  and  square  respectively. 
Show  that  the  sum  of  the  areas  of  these  figures  will  be  least 
when  the  wire  is  cut  in  the  ratio  tt  :  4. 


CHAPTER   VII 

TOEMULAS  POK  DIFFERENTIATION  OF  LOGARITHMIC  AND 
EXPONENTIAL  FUNCTIONS.    HYPERBOLIC  FUNCTIONS 

75.  The  formulas  of  Art.  52  are  not  sufficient  to  enable  us 
to  determine  the  derivative  of  any  other  than  an  algebraic 
function,  as  can  readily  be  seen  by  an  example. 

Example.     Suppose  that  it  is  required  to  find  -^,  if 

dx 

y=x2logw(x3  +  l). 
dy  =  a«ilogy(*  +  l)l+  jlogio(x3  +  1} jd*  by  Formula  IV) 

(XX  CtX  ctx 

=  ^?lQgio(^8  +  1)?  +  2  x  logl0  (a*  +  1),  by  Formula  VII. 

(JjJb 

Since  none  of  the  formulas  of  Art.  52  gives  a  means  of 

finding  ^2A — X_J;  JL  is  not  determined. 

ctx  ux 

76.  The  following  formulas  with  those  of  Art.  52  will  enable 
us  to  find  the  derivative  of  any  logarithmic  or  exponential 
function.  ,,  -,   7 

VIIL  <nog„M=ig»<log^ 

dx  u  dx 

j^-     d  loge  u  _1  du 
dx  u  dx 

v  dau       udu1 

X.  - —  =  au  —  loge  a. 

(JjtJU  LhJU 

•TTT  \AJ\J  «/  \JjVV 

LvJb  \AjJU 

XII.  —  =  vu"-1—  +  u*— loge  ^. 

dec  dx  dx 

79 


80 


DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


In  these  formulas,  u  and  v  are  functions  of  x;  a  is  any 
constant  capable  of  being  a  base  to  a  system  of  logarithms; 
and  e  is  a  particular  constant  whose  value  to  seven  places  of 
decimals  is  calculated  further  on  and  found  to  be  2.7182818  •••. 


77.   Derivation  of  formulas. 

Proof  of  VIII. 

Let 

y  =  loga  w. 

Let  x  =  x0. 

.-.  Wo  =  logaw0. 

Let  x  —  x0  -+-  Ace. 

.-.  y0  +  Ay  =  loga(u0  +  Au). 

Subtract. 

•'•  &y=loga(u0  +  Au)- 

-,        iu  4-  Ait 

=  loga-^= 

w0 

-lOga^0 

Divide  by  Ace. 

a       log.(l  +  ^ 

Ace                Ace 

1     Aw     uQ     i 

w0    Ace    Aw 


A?tN 
w0 

identically, 


1     Aw    i 
-_.        .logo   1  + 


.  dy 

dx 


limit 
Ax=0 


'Ay' 

Ace 


a»\2 

w0    Ace  V         wo/ 

since  m  loga  iV  =  loga  ^r 


1  ^  limit 
17  '  Ax  =  0 


1   dw 

Wq  cix 


Aw 
Ace 


limit 
Ax=0 


"S,+Sr 


limit 
*  Ax  =  0 


i0.,  .i+^hU 

w0 


It  remains  to  find 


limit 
Ax=0 


L 


LOGARITHMIC  AND  EXPONENTIAL   FUNCTIONS     81 


Let  ^-  =  z.     As  Ax  =  0,  Au  =  0,  and  — ,  or  z,  =  go. 
Au  Au 


limit 
*  '    Az=0 


"log;(l+^" 


limit 


loga[l  +  - 


To   find   lirait 


log„fl+- 


expand  [1  +  -]     by   the   Bi 


nomial  Theorem,  assuming  z  to  be  a  large  positive  integer. 


l+±)  =l+z 


l+^y+!fc^y+^gj 


12 


|3 


Now 


limit 

2  =  00 


limit 

Z  =  cc 


1  +  ~z 


i_i  ri-iYi-2 


1+1+ 


+ 


+ 


=  the  infinite  series  l  +  l+r^  +  r^+  '"• 


Call  the  value  of  this  series  e. 


dy 

dx 


1  du 


log.  e. 


(2  log.  u     1  c?m  , 

los„  e. 


dx 


u  dx 


Note.     In  showing  that    ™l  |log«  ( 1  -f  -  J      =  logae,  a  number  of 

assumptions  were  made  that  would  require  closer  investigation  for  a 
rigorous  proof.  For  a  rigorous  proof  the  student  is  referred  to  Byerly's 
Differential  Calculus,  pages  50,  51,  and  52. 

G 


82 


DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


To  calculate*  the  value  of  e  from  the  infinite  series,  to  seven 
decimal  places : 

0      1,1,1,1,1,1,1,1,1,1,    1    .    1 

=        1.0 

+  1.0 
+  0.5 

+  0.166666666 
+  0.041666666 
+  0.008333333 
+  0.001388888 
+  0.000198412 
+  0.000024801 
+  0.000002755 
+  0.000000275 
+  0.000000025 
=    2.7182818  ... 

Proof  of  IX. 

This  is  a  special  case  of  VIII,  in  which  a  =  e.     In  this  case 

d  log,  u      1  du     •        -,  -, 
^—  = ,  since  loge  e  =  1. 

Logarithms  to  the  base  e  are  called  natural,  or  sometimes 
hyperbolic,  logarithms.  When  the  base  of  a  logarithm  is  not 
given,  it  is  assumed  that  the  logarithm  is  to  the  base  e. 


Proof  of  X. 

Let 


y  =  a 


Take  the  logarithm  of  each  member  of  the  equation. 
.-.  loge2/  =  logeaM 
=  u  loge  a, 


LOGARITHMIC  AND  EXPONENTIAL  FUNCTIONS     83 

Differentiate  each  member  of  the  equation  with  respect  to  x. 

...  !^  =  ^logea. 

y  dx      dx 

dy        du  -i 
.-.  -f-  =  y  —  \ogea 
dx        dx 

=  au  -¥■  loge  a. 
dx 

Proof  of  XI. 

This  is  a  special  case  of  X,  in  which  a  =  e.     In  this  case 

d^  =  eu<h±j  since  l0gee=l. 
dx  dx 

Proof  of  XII. 

Let  y  =  uv. 

Take  the  logarithm  of  each  member  of  the  equation. 
.-.  loge  y  =  v  loge  u. 

Differentiate  each  member  of  the  equation  with  respect  to  x. 

1  dy      v  du  .  dv  -, 
.-.  --JL  =  -—  +—  l0geU. 


u 


y  dx      u  dx     dx 

dy        fv du  ,  dv-,   ^ 
dx        \u  dx      dx 


=  vuv~^  —  -\-uv—  loge  lh  since  y  =  uv. 

LLJb  LiiJb 


This  formula  can  be  easily  remembered  by  noting  that  it 
can  be  derived  by  differentiating  uv,  first,  as  if  v  were  constant 
and  it  variable ;  second,  as  if  u  were  constant  and  v  variable ; 
and  then  adding  the  two  results. 


84         DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


EXERCISES 


dy 


Find  -^  in  each  of  the  following  cases : 
ax 

1.    y  =  log(a?  +  2x). 


2.  y  =  log  VaJ  — 1. 

3.  y  =  log  (a?  + 1)2. 

4.  2/  =  cc2loga;. 

5.  2/  =  x  1°§  (a*2  +  !)■ 


a2  +  2z 


2(z-l) 

Ans. 

Ans.   x  (1  -f-  2  log  #). 
^4ns.  -l^T  +  log(aj2  +  l). 


a^  +  l 


6.  2/  =  log  (^  + 1) Vl  —  x2. 

7.  2/  =  ea:logcc. 

8.  y  —  xnax. 

9.  2/  =  #2e3a:. 


Ans. 


3x3  —  x 
x4-l  ' 


Ans.  — \-ex  log  x. 


x 


Ans.  xn~1ax  (x  log  a  +  n). 
Ans.  a;eSx(2  +  3  a). 


10.   2/  =  e4*  log  Vl  +  x2.  Ans.  e4* 


11.   2/  =  log  (ex  +  e~x). 


21og(l  +  z2)  + 


x 


L 


^.TIS. 


1  +  Z2 

e2*-l 
e2*  +  l' 


12.    2/  =  logio(^2  +  5^). 

^4WS.  m 20x  +  5,  where  Jf  =  log10e  =  .43429  •••. 

x2  -\-5x 


13.    y  —  5xi+Sx. 


Ans.  5**+3x(2x  +  3)log5. 


14.   y  =  2x*  —3x*  +6x6  —  61og (x5  +  1).  ^4ns.  — -• 

#2  -j-  x* 


15.  2/  =  ^. 

16.  2/  =  ee*- 


1-2 


Ans.  of    (1  —  logic). 
Ans.  exeeX. 


HYPERBOLIC  FUNCTIONS  85 


ex  —  e~x  A  4e2* 

17.    ?/  = Aiis. 


ex  +  e~x  (e2x  -f-  1)' 


i 


18.  y  =  a}°*x.  Ans.  0. 

clu    fly 
In  each  of  the   two  following   exercises,  find  -^,  — ,  and 
7  cZz    cZa? 

then  -^: 
dx 

19.  ?/ =  (1  +  e*  +  e2*),  «  =  log(x2  -2x  +  3).  - 

^w«.  dy  =  2(x-l)(2x2-4.x  +  7). 

dx  J 

20.  ?/  =  log  (2  2  -  z2),  z  =  e~\  Ans.  %L  =  2(1-0. 

do?        2  —  ex_1 

21.  In  the  curve  2/  =  log  (x2  +  2x),  find  the  abscissas  of  the 
points  at  which  x  and  y  change  at  the  same  rate. 

Ans.  x=  ±  V2. 


- — -  +  log\/    -~      for  maximum 
x  -\- 1  *x°  —  l 

La  minimum  values. 

^4?*s.  Max.  x  —  —  1.     Min.  a;  =  + 1. 


HYPERBOLIC   FUNCTIONS 


!U  s>—U 


78.   Definitions.     The  expression  is  denoted  by  sinh  u 

and  read,  "  hyperbolic  sine  of  uP 

eu  -4-  6~u 
The  expression   — X is  denoted  by  cosh  u  and   read, 

"  hyperbolic  cosine  of  u."     Also, 

smh  u        €>u 6~ u 

— : — ,  or ,  is  denoted  by  tanh  u.     Eead,  "hyperbolic 

cosh  u        eu  +  e~u  '       Jr 

tangent  of  u." 

■    Cosh  11  6U  -\~  6~u 

-r-, — ,  or  — !- ,  is  denoted  by  coth  u.     Eead,  "hyperbolic 

smh  u        eu  —  e~w  J  r 

cotangent  of  u." 


86  DIFFERENTIAL  AND  INTEGRAL    CALCULUS 

1 


sinh  it 
of  u." 
1 


is  denoted  by  cosech  u.     Bead,  "  hyperbolic  cosecant 
is  denoted  by  sech  u.    Read,  "hyperbolic  secant  of  u." 


cosh  u 

EXERCISES 

1.  Establish  the  following  identities: 

cosh2  u  —  sinh2  u  =  1.  tanh2  u  -f-  sech2  u  =  1. 

coth2  u  —  cosech2  u  =  1.  sinh  2u  —  2  sinh  u  cosh  u. 

sinh  (u  ±v)  =  sinh  u  cosh  v  ±  cosh  m  sinh  v. 

2.  If  %  is  a  function  of  x,  prove  that : 


d  sinh  it           -.      c?w 
=  cosh  u  — 

dx                     dx 

d  cosh  u       •  -,      du 

=  smn  u  — 

dx                      dx 

d  tanh  u           ■,  2     du 
=  seclr  u  — 

ax                       ax 

d  coth  u                  -.  2 
=  —  coseclr 

dx 

du 

u  — • 

dx 

d  cosech  u  _ 
dx 

cosech  u  coth  u  — 
dx 

=  —  sech  u 

dx 

tanh  it  — 
dx 

Definitions.     If  u  =  sinh  y,  then  y=  sinh-1  u. 
Similarly  for  the  other  hyperbolic  functions. 

3.    If  u  is  a  function  of  x,  prove  that : 

du 
d  sinh-1  u  dx 


dx 

-\A^  +  1 

Proof.     Let  y  =  sinh-1  u. 

.-.  sinh  y  =  u. 

die 

du 

t      dy     du 
.-.  cosh?/  —  =  — • 
dx     dx 

dy        dx 

dx  ~~  cosh  y 

dx 

Vu2  +  1 

du 

.-.  d  sinh- 

1  u          dx 

dx  V%2  + 1 


HYPERBOLIC  FUNCTIONS  87 

4.    If  u  is  a  function  of  x,  prove  that : 

du  du 

d  cosh-1  u  dx  d  tanh-1  u        dx 


dx  -^/u2  _i  dx  1  —  u2 

du  du 

d  coth-1  u        dx  d  cosech-1  u  dx 


dx  1  —  u2  dx  ftVl  +  ft2 


du 
d  sech-1  u  dx 


d®  u  Vl  —  u2 

5.    Show  that: 


sinh"1  u  =  log  (u  +  Vft2  + 1).  cosh"1  u  =  log  (u+  VuF^l). 

tanh"1  u  =  \  log  1±1*.  coth-1  u  =  ±  log  ^±1 

2        1-u  2     Bu-1 


lM==wl±^±i2.  fipnW,_l.      1+Vl- 


cosech-1  u  =  log     m  VJ-t-^t  sech"1  w  =  log 


ft' 


ft  &  ft 


CHAPTER   VIII 

FORMULAS  FOR  DIFFERENTIATION  OF  TRIGONOMETRIC 
AND  ANTI-TRIGONOMETRIC  FUNCTIONS 

79.  The  following  formulas  with  those  of  Art.  52  will  enable 
us  to  find  the  derivative  of  any  trigonometric  or  anti-trigono- 
metric function. 

VTTT          d  sin  u  du 

.aJ.11.        — =  cos  u  — 

dx  dx 

VTTT         d  cos  u  •       du 

XIV.         ■ =  -SlllM — 

dx  dx 

VTT         d  tan  u  2    du 

XV        =  secJ  u  — 

dx  dx 

_„,x      .   d  cot  u  o    du 

XVI.        ■ =  —  cosec2  u — 

dx  dx 

__XTTT         d  sec  u  .         du 

XVII.        =  secwtanw— • 

dx  dx 

XVIII.    d  C0SeC  '( =  -  oosec  u  cot  u  r^. 
dx  ax 

XIX.      dTerBM  =  sin«^. 

dx  dx 


XX. 


XXI. 


XXII. 


du 
d  sin"1  u  dx 


dx  Vl  —  u2 

du 
d  cos-1  u  dx 


dx  Vl  —  u2 

du 
dtan-1^     dx 
dx  1  +  u2 

88 


TRIGONOMETRIC,  ANTI-TRIGONOMETRIC  FUNCTIONS     89 


XXIII. 


XXIV. 


XXV. 


XXVI. 


du 
d  cot-1  u  dx 


dx 

1  +  ir 

d  sec-1 u 

du 

dx 

dx 

u  Vtt2  —  1 

d cosec-1  u 

— 

du 
dx 

dx 

u  Vw2  —  1 

du 

d  vers-1  u 

dx 

dx  V2  u  —  u2 


80.   Derivation  of  some  of  the  formulas. 
Proof  of  XIII. 

Let  y  =  sin  u. 

Let  x  =  x0.  .*.  y0  =  sin  u0. 

Let  x  =  x0  +  Ax.     .-.  2/0  +  A?/ =  sin  (w0  +  Aw). 
Subtract.  .-.  Ay=  sin  (u0  +  Au)  —  sin  u0. 

In  the  trigonometric  formula, 

sin  A  —  sin  B  =  2  cos  1  (A  +  5)  sin  J  (-4  —  -B), 
let  A  =  u0  +  Au,  and  5  =  w0. 

.-.  sin  (u0  +  Aw)  —  sin  uQ  =  2  cos  |(2«0  +  Aw)  sin i  Aw. 
.  •.  Ay  =2  cos  |(2  m0  -f  Aw)  sin  \  Au 

=  2  cos  (ii0  +  — ^  ]  sin  \  Aw. 

Divide  by  An    .-.  ^L  =  2  cos  („,  +  ^Vin»Aw 
J  Az  \,  2  J      Ax 

=  2coSf«0  +  A^V_Hil^     An 


2  J      Au  Ax 


=  cog[«0  +  AJfViniAw.A». 

2/    {4»        A* 


90 


DIFFERENTIAL   AND  INTEGRAL   CALCULUS 


Pass  to  limits. 


dy 

dx 


limit 
Ax=0 


=  COS  Uq 


,  Au 
COS  (  u0  +  — 


limit 
Ax=0 


"sjn  1  Ait 

1^ 


limit 
Ax  =  0 


"Ait" 
A& 


dit 


limit 
A£  =  0 


"sin  ^  A?/ 


It  remains  to  find   .A 

Ax=0 


'sin  -i  Ait" 

L     1A^     . 


As  Ax  =  0,  Au  =  0.    .-.  J™1* 


Let  ^  Aw  =  0. 


limit 
'  '  Ait=0 


"sin  \  Au 


"sin  l  Ait" 
l   iAu 


limit 
Aw  =  0 


limit 
0  =  0 


"sin  i  Au" 
.    i  Am 


sin  0 


With  0  (Fig.  28)  as  center  and  any  distance  r  as  radius, 

describe  a  circle.  Take  any  angle  0 
expressed  in  circular  measure.  From 
A  draw  a  perpendicular  to  OA  to 
meet  O-S  produced  in  D. 

Arc  J.5  =  r6,  AD  =  r  tan  0,  and 
5(7  =  r  sin  0. 

Now  by  geometry, 

CB  <  arc  .45  <  AD. 

r  sin  0  <  ?*0  <  r  tan  0. 
\  sin  (9  <  0  <  tan  (9. 

tan0 


Fig.  28. 


Divide  by  r. 
Divide  by  sin  0. 


.'.  K  - 
.-.  K 


sin  0       sin  0 
0 


sin  0 


<  sec  0. 


Now  as  0  =  0,  sec  0  =  1. 


sin  0 


is  always  greater  than  1 


and  less  than  something  that  approaches  1  as  its  limit. 


TRIGONOMETRIC,  ANTI-TRIGONOMETRIC  FUNCTIONS     91 


Therefore  W* 
0  =  0 


sin  6' 
d  sin  u 


=  1. 


limit 

Ax  =  0 


"sin  i  Am 


=  1. 


=  cos  w 


Proof  of  XIV. 

Let  y  =  cos  u. 


l[2~U 


dy 


f  i 


Tx  =  C0S[2-U 


d[-  —  u 


dx 
du 


by  Formula  XIII, 


d  cos  u  .        du 
=  —  sin  u  —  • 

LttJO  (JUL/ 


Proof  of  XV. 
Let 


y  =  tan  u. 
sin  u 


y 


cos  u 


d  sin  u       •        d  cos  u 

cos  m sm  u 

dy  dx  dx 

dx  cos2  u 

_  cos2  u  4-  sin2  w  cZm 


,  by  Formula  VI, 


'  u         dx 

—  eon" 


1      du  o     du 

— =  secJ  u  — • 

cos"  u  dx  dx 


d  tan  u 


—  oor>2 


dx 


sec"  u 


du 
dx 


Proof  of  XX. 
Let 


y  =  sin-1  u. 


,\  sm  y  =  u. 


92  DIFFERENTIAL  AND  INTEGRAL    CALCULUS 

Differentiate  each,  member  of  the  equation  with  respect  to  x. 


du     du 
cos  y  -£=  — 
dx     dx 

du  du 

dy        dx  dx 

dx  ~~  cos  u  ~  -y/i  _u2 

du 
d  sin-1  u  dx 


dx  -y/J  _ 


v: 


The  proofs  of  the  other  formulas  are  left  as  exercises  to  the 
student. 

EXERCISES 

1.    Prove  the  following : 

du 

d  cot  u  „     du  d  cot-1  u  dx 

=  —  cosec-  u  — 


dx  dx  dx  1  +  u2 

du 

d  cosec  u  ,      du      d  cosec-1  u  dx 

=  —  cosec  u  cot  u  — 


dx  dx  dx  u^/u2  —  1 

du 

d  cos-1  u  dx 


dx  yX 


u 


2.    Prove  the  following : 

du 

d  sec  u  du  d  sec-1  u  dx 

=  sec  u  tan  u  — • 


dx  dx  dx  %Vw2  —  1 

du 

dversw        .       du  dvers-1u  dx 

=  sm« 


— i Z>L1±  U>  -rr~  •  7  —         ,- j 

dx                   dx                                   dx  V2  u  —  w2 

d  tan-1  u _     dx 
cte        ~  1  +  u2 


TRIGONOMETRIC,  ANTI-TRIGONOMETRIC  FUNC1  IONS     93 

Find  —  in  each  of  the  following  cases  : 
dx 

3 .    y  =  sin  3  x  cos  2  x.     yl/>.s.  3  cos  2  aj  cos  3x  —  2  sin 2  a;  sin  3  a?. 


4.  _y  =  sin2 2  cos  2  a?. 

5.  y  =  sin2  as  cos2  a?. 

6.  ?/ =  tair  5:/;. 

7.  y  =  log  sec2  x. 

8.  y  =  log  tan2  3  a?. 

9.  r/  =  log  cos  4  a;. 

10.  y=logtanf|  +  |\ 

11.  y  =  e"*  log  sin  nx. 

12.  3/  =  ef*  sin-1  a?. 

13.  v^tan-1- LJ. 

J  2 

14.  v  =  COS-1  - — . 

lo.  y=  sec  x 

1-f-z2 

16.  y  =  sin-1  Vsin  x. 

17.  gf  =  tarr 


cr  -  3  "x~ 


^<bi&  (1  —  4  sin2  a?)  sin  2  #. 
,4ns.  |  sin4aj. 

Ans.  10  tan2  5  a;  sec2  5  a;. 

^L/t.5.  2  tan  a:. 

-4/is.  6  sec2  3  a;  cot  3  a?. 

-4><s.   —  4tan4ic. 

^./t.s.  sec  a?. 

-4/t.s-.  eax  [/'.  cot  wa  +  «  log  sin  ?ia;]. 
1 


yl/t.s.  e8* 


.Vl-z2 


4-  a  sin-1  # 


-4/i.s. 


2 


-2 


As. 


K  V3  a;4  -  2  ar>  -  1 


18.  y  =  sec  a;  4-  low  tan  — 

-  _  2  sin-1  #  ,  ,      1  —  aj 

19.  y=  +logT- — 

Vl  -  z2  1  +  * 

2 


Ans.  -J  Vl  +  cosec  a;. 

.  3  a 

or  4-  ar 

-4ft.s.  sec2  x  cosec  a;. 


-4ws. 


2  a;  sin-1  aj 


20.  y  =  1  a;Va2  -  a^  4-  -  sin"1  -• 

"  2  a 

01  j.     _i2 a;  —  a  ,  .       ,2ft —  a? 

21.  y  =  tan  : f-  tan-1         _  ■ 

c/V3  ars/3 


-4ns.   V"2  —  a?2. 
As.    0. 


94         DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

00      ,      v 1  +  a*      i  l      1  +  x  +  x2  ,     /o  .     _i  #V3 

22.   y  =  log  — I—  +  ^log ,  ;       +V3tan1- -• 

1  —  x  1  —  x  -\-  xz  1  —  ar 

6 


^4?is. 


1  —  a?6 

23.    In  the  curve  y  =  log  sec  x,  find  the  abscissas  of  the  points 
at  which  x  and  y  change  at  the  same  rate.         Ans.  (An  + 1)- • 


24.  A  vertical  wheel  of  radius  8  feet  makes  40  revolutions 
per  minute  in  the  positive  direction  of  rotation,  about  a  fixed 
axis.  Find,  in  feet  per  minute,  the  horizontal  and  vertical 
speeds  of  a  point  on  the  circumference  (a)  30°  from  the  hori- 
zontal ;  (5)  at  the  highest  point  of  the  circumference ;  (c)  180° 
from  the  horizontal ;  (d)  at  the  lowest  point  of  the  circum- 
ference. 

Ans.  Horiz.  vel. :  (a)  -320tt;  (6)  -640tt;  (c)  0 ;  (d)  640  tt. 
Vert,  vel.:  (a)  320V3tt;  (5)  0;  (c)  -640tt;  (d)  0. 

25.  A  vertical  wheel  of  radius  8  feet  is  moving  about  a  fixed 
axis.  If  the  horizontal  and  vertical  speeds  of  a  point  on  the 
circumference,  at  a  given  instant,  are  20  feet  per  second  and 
10  feet  per  second,  respectively,  find  the  angle  through  which 
the  wheel  turns  in  1  second.  Ans.  160.1°. 

26.  A  carriage  wheel  rolls  along  a  level  road,  the  axle 
moving  at  the  rate  of  8  miles  per  hour.  Find,  in  feet  per 
second,  the  horizontal  and  vertical  speeds  of  the  point  of  the 
circumference  originally  on  the  ground  when  the  wheel  has 
rolled  through  an  angle  of  (a)  30° ;  (6)  60°;  (c)90°;  (d)  180°; 
(e)  270°;  (/)  360°. 

Suggestion.     First  derive  the  equations  of  the  cycloid 

x  =  a  {6  —  sin  0),  y  =  a(l  -  cos  0). 

Ans.    Horiz.  vel.  (a)  1.57;    (b)  5.87;    (c)  11.73;    (d)  23.47; 
(e)  11.73;  (/)  0. 
Vert.  vel.  (a)  5.87;  (6)  10.16;  (c)  11.73;  (d)  0;  («)  -11.73; 

(/)  o. 


TRIGONOMETRIC,  ANTI-TRIGONOMETRIC  FUNCTIONS     95 

27.  The  crank  and  connecting  rod  of  a  steam  engine  are  2  and 
6  feet  in  length  respectively.  If  the  crank  revolves  uniformly 
at  the  rate  of  3  turns  per  second,  what  is  the  rate,  in  feet  per 
second,  at  which  the  piston  is  moving  when  the  angle  which 
the  crank  makes  with  the  horizontal  is  (a)  0° ;  (6)  45° ;  (c)  90° ; 
(d)135°;  (e)  180°;  (/)  270°. 

Ans.  (a)  0;  (b)  10.54  tt;  (c)  12  tt;  (d)  6.43  tt;  (e)  0; 
(/)    -12  IT. 

Determine  the  maximum  and  minimum  points  and  points  of 
inflexion  of  the  following  curves.     Plot  the  curves. 


28.  y  =  sin  x. 
ns.   Max.  (4n 

29.  y  =  cosx. 


Ans.   Max.  (4  n  + 1)  ^  •     Min.  (4  n  +  3)  £  ■     Pts.  of  inflex.  wtt. 

Li  Li 


TT 


Ans.    Max.  2  wtt.     Min.  (2  w  +  1)tt.     Pts.  of  inflex.  (2  n  + 1)  ^ 

Li 

30.    i/ =  sin  2  a; -f  cos  2  a?. 

J.ws.   Max.  (8  w  + 1)  I  •     Min.  (8  n  +  5)  £  • 

o  8 


TT 


Pts.  of  inflex.  (4n  +  3)£ 

31 .  y  =  sin  &  —  cos  x.  ° 

-4ns.   Max.  (8  n  +  3)  j  •     Min.  (8  w  +  7)  -  • 

Pts.  of  inflex.  (4n  +  l)-« 

y4 

Determine  the  maximum  and  minimum  values  of  r  in  each 
of  the  two  following  equations.     In  each  case,  plot  the  curve. 

32.  r  =  a(sin2  0  +  cos2  0). 

33.  r  =  a(sin#  —  cos0). 

34.  An  open  gutter  with  cross  section  an  arc  of  a  circle  is 

to  be  bent  from  a  piece  of  tin  12  inches  wide.     Find  the  width 

across  the  top  when  the  carrying  capacity  of  the  gutter  is  a 

maximum.  24 

Ans.    — 

TT 


CHAPTER   IX 


EOLLE'S  THEOEEM.    LAW  OF  THE  MEAN. 
TAYLOE'S  THEOEEM 

81.  Before  proceeding  to  a  discussion  of  the  theorems  named 
above,  we  shall  prove  the  following  theorem : 

In  the  equation  y=f(x),  where  f(x)  is  a  single  valued  and 

continuous  function  of  x,  as  x  increases  through  x0,  if  f(x)  is 

dy 
increasing,  —         is  positive  or  zero,  and  if  f(x)  is  decreasing, 

dy  .  ° 

-~         is  negative  or  zero. 


Proof. 

Let  x  =  x0. 

Let  x  =  x0  +  Ace. 


2fo  +  Ay=/(a>0  +  Aa;). 

.    A?/      /(aJb  +  Aaj)-/(a\)) 


AflJ 


A# 


Suppose  that  f(x)  is  increasing  as  a  increases  through  x0. 
Under  this   supposition,  f(x0  +  Ax)>f(x0),   and    therefore 
/(^0  +  Ax)—f(x0)  or  A?/  is  positive.     Now  Aa  is  positive,  since 


limit 


Ay 

Ax 


or 


dy 


dx 


by  supposition  a  increases.      Therefore  Ax±q 
is  positive  or  zero. 

Suppose  that  f(x)  is  decreasing  as  x  increases  through  x0. 

Under  this  supposition,  f(x0  +  Aa?)  <  /(a?0),  or  Ay  is  negative. 


A?/ 
Therefore  — -  is  negative. 

Ax  & 

negative  or  zero. 


Therefore 


96 


limit 
A#  =  0 


Ay 
Ax 


01 


n  dy 


dx 


IS 


ROLLERS   THEOREM  97 

82.  If  between  two  values  of  x,  namely  x  =  a  and  x  =  b,  a 
function,  f(x),  is  single  valued  and  continuous  and  has  a  con- 
tinuous derivative,  and  if  f(a)  =  fib)  =  0,  then  there  is  at 

dv 
least  one  value  x0  of  a  between  a  and  6  at  which  -f-  is  zero. 

This  is  known  as  Rolle's  Theorem.  The  geometrical  state- 
ment is  as  follows : 

If  the  curve  whose  equation  is  y  =  f(x),  where  fix)  is  single 
valued  and  continuous  for  all  values  of  x  between  two  values 
a  and  b,  crosses  the  a>axis  when  x  =  a  and  x  =  b  and  does  not 
have  a  cusp  for  any  value  of  x  between  a  and  b,  the  tangent 
line  to  the  curve  will  be  parallel  to  the  x-axis  for  at  least  one 
value  of  x  between  a  and  b. 

Proof  of  Rolle's  Theorem. 

Since  f(x)  is  zero  when  x  =  a  and  x  =  b,  and  is  continuous, 
it  must  have : 

Case  I  been  zero  for  all  values  of  x  between  a  and  b,  or 
Case  II  increased  from  zero  and  finally  decreased  back  to 

zero,  or 

Case  III  decreased  from  zero  and.  finally  increased  back  to 

zero. 

Case  I.  If  f(x)  is  zero  for  all  values  of  x  between  a  and  b, 
its  derivative,  — ,  is  zero,  not  only  for  one  value,  but  for  all 
values  of  x  between  a  and  b. 

Case  II.  If  fix)  increased  from  zero  and  finally  decreased 
back  to  zero,  it  must  have  changed  from  increasing  to  decreas- 
ing at  least  once.     Therefore,  by  Art.  81,  its  derivative,  ^, 

dx 

must  have  changed  from  positive  to  negative  at  least  once. 
Now  a  continuous  function  can  change  in  sign  only  by  passing 

through  the  value  zero.     Therefore  —   must  have  been  zero 

dx 

for  at  least  one  value  of  x  between  a  and  b. 

The  Proof  of  Case  III  is  entirely  similar  to  that  of  Case  II. 


98  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

83.  As  an  illustration,  consider  the  equation 

y  =  x4-8x?  +  22x2-24:X. 

The  function  y  is  zero  when  x  =  0  and  x  =  4t.  Then,  by 
Rolle's  Theorem,  the  derivative  must  be  zero  for  at  least  one 
value  of  x  between  0  and  4.     That  such  is  the  case  can  readily 

be  seen.  flll 

^  =  4ar3-24^  +  44a-24 
dx 

=  4(x-l)(x-2)(x-3); 

and  an  examination  of.  this  function  shows  that  it  vanishes 
three  times  between  0  and  4,  namely,  at  1,  2,  and  3. 

EXERCISES 

1.    Give  a  geometrical  proof  of  Rolle's  Theorem. 

In  each  of  the  following  equations,  show  that  —  is  zero  for 

dx 

at  least  one  value  of  x  between  each  two  for  which  y  is  zero. 
In  each  case,  plot  the  curve. 

2.  y  =  xs  +  6  x2  +  11  x  +  6. 

3.  y  =  a?-12x  +  ll. 

4.  y  =  x*-3xs  +  x2  +  3x-2. 

5.  y  =  x3  —  x2  —  16  x  +  16. 

84.  If  between  two  values  of  x,  namely  x  =  a  and  x  =  b,  a 
function,  f(x),  is  single  valued  and  continuous  and  has  a  con- 
tinuous derivative,  then  there  must  be  at  least  one  value  xx  of 
x  between  a  and  b  at  which 

b  —  a  J  x   J 

This  is  known  as  the  Law  of  the  Mean.  It  is  readily  seen 
that   Rolle's  Theorem   is   but   a   special   case  of   it,  because 

if  /O)  =  f(P)  =  0,    f^Zi^  =  /'Oi)>   a<xi<  h   becomes 
0=f'(xl),  a<x1<b>  which  is  Rolle's  Theorem. 


LAW  OF  THE  MEAN 


99 


85.  The  Law  of  the  Mean  has  a  simple  geometrical  inter- 
pretation. Plot  the  curve  y=f(x)  (see  Fig.  29).  The  left- 
hand  member,     ^  ;  ~^a\  is  the  slope  of  the  chord  AB.     The 

b  —  a 

right-hand  member,  f'(xi),  ct  <  xx  <  b,  is  the  slope  of  the  tan- 
gent line  to  the  curve  at  a  point  on  the  curve  for  which  x 
has  a  value  between  a  and 
b.     The  Law  of  the  Mean 
stated  geometrically  is  there- 
fore: 

In  the  curve  y  =  f(x)  in 
which  f(x)  is  single  valued 
and  continuous  and  has  a 
continuous  derivative  for  all 
values  of  x  between  a  and 
b,  the  slope  of  the  chord  joining  the  points  for  which  x  =  a 
and  x  =  b  is  equal  to  the  slope  of  the  tangent  line  at  some 
point  for  which  x  has  a  value  between  a  and  b. 

86.  To  establish  the  Law  of  the  Mean,  we  proceed  as 
follows : 

From  f(x)  we  build  up  a  function,  F(x),  which  is  zero  when 
x  =  a  and  x  =  b.  Then,  by  applying  Rolle's  Theorem  to  F(x), 
we  are  able  to  obtain  the  desired  result. 


Fig.  29. 


Proof  of  the  Law  of  the  Mean. 
Let  m-.M 


=  K. 


b  —  a 

Since  a  and  b  are  constants,  iTis  constant. 
Clear  of  fractions  and  transpose. 

.'.f(b)-f(a)-(b-a)K=0. 

From  f(x)  build  up  the  function 

F(x)=f(x)-f(a)-(x-a)K, 
where  if  has  the  value  given  it  in  (1). 


a) 


(2) 


100       DIFFERENTIAL  AND  INTEGRAL    CALCULUS 

Then  F(a)=f(ci)-f(a)-(a-a)K=0.  ■ 

Also,  F(b)  =/(&)  -f(a)  -(b-a)K 

=  0  because  of  (2). 
Differentiate  F(x),  remembering  that  f(a)  and  K  are  con- 
stants'  .-.  F\x)=f\x)-K. 

Since  F(a)  =  F(b)  =  0,  .-.  F'fa)  =  f(xx)  - K=  0,a<x1<b, 
by  Eolle's  Theorem. 

.-.  K=f'(x1),   a<x1<b. 
Substitute  in  (1). 

■■■mbZ{(a) =/>■),  «<*,<&■ 

This  equation  is  frequently  written  in  the  form 
f(b)=f(a)  +  (b-a)f'(x1),   a<x,<b. 

87.  If  between  two  values  of  x,  namely  x  =  a  and  x  =  b,  a 
function,  /(cc),  is  single  valued  and  continuous  and  has  a  first 
and  second  derivative  each  of  which  is  continuous,  then 

f(b)  =f(a)  +  (b  -  a)f'(a)  +  (&  ~  ^ /"&),   a<x,<b. 
Proof. 

/(»)-/(«)- (ft -q)/'(q)  =  g 

Since  a  and  6  are  constants,  iT  is  a  constant. 
Clear  of  fractions  and  transpose. 

...  f(b)  -/(a)  -  (6  -  «)/'(«)  -  ^=^  A'=  0.  (2) 


From  f(x)  build  up  the  function 

F(x)  =f(x)  -/(a)  -  (x-  a)f'(a)  -&=^K, 
where  iThas  the  value  given  it  in  (1). 


LAW  OF  THE  MEAN  101 

Then  F(a)  =/(«)  -/(a)  -  (a  -  a)f'(a)  -  ^^  K=  0. 

Also,  F(b)  =/(&)  -/(a)  -  (6  -  a)/'(a)  -  ^plV 
=  0  because  of  (2). 

Differentiate  F(x),  remembering  that  /(a),  /'(a),  and  ^  are 

constants. 

.-.  F<(x)=f\x)-f<(a)-(x-a)K. 

Since  F(a)  =  F(b)  =  0, 

...  FXh)=fVh)-f'(a)-(ht-a)K=0}   a<lh<b,     (3) 

by  Bolle's  Theorem. 
Differentiate  .F'(a;). 

...  F"(x)=f"(x)-K. 

Now  i^'(a)  =/'(a)  -/'(a)  -  (a  -  a)/r=  0.     Also,  F'fa)  =  0 
because  of  (3). 

.-.  F"(h2)=f"(h2)-K=0,   a<h2<h1} 

by  B-olle's  Theorem. 

.-.  K=f"(h2),   a<h2<7h. 

Since  h2  <  hx  <b,  .-.  7i2  <  b. 

.-.  K=f"(h2),   a<h2<b. 
Substitute  in  (1). 

■  •  (&  _  ca2 —  =f  Qh),  a<h2<b. 

"IT" 

Call  ft2  =  a^.     Clear  of  fractions  and  transpose. 


•••  /(&)  =/(«)  +  (6  -  «)/'(«)  +  ^=^-7"K>,  a  <  4  <  6. 


102       DIFFERENTIAL  AND  INTEGRAL    CALCULUS 

EXERCISES 

1.  Find,  by  the  Law  of  the  Mean,  a  point  on  the  curve  y  =  x? 
at  which  the  tangent  line  to  the  curve  is  parallel  to  the  line 
joining  the  points  (2,  8)  and  (5,  125).         Ans.    (Vl3,  13V13). 

2.  Using  the  Law  of  the  Mean,  find  the  equation  of  the  tan- 
gent line  to  the  curve  y2  =  4:X,  which  is  parallel  to  the  line 
joining  the  points  (0,  0)  and  (4,  4).  Ans.  x  —  y  + 1  =  0. 

3.  Using  the  Law  of  the  Mean,  find  the  equation  of  the  tan- 
gent line  to  the  curve  y2  =  8  x,  which  is  parallel  to  the  line 
2x-3y  +  8  =  0.  Ans.   2x-3y  +  9  =  Q. 

4.  Show  that  F(x)  in  Art.  86  represents  the  length  QP 
where  P  and  Q  are  points  on  the  curve  y=f(x)  and  the  chord 
AB  respectively  having  the  same  abscissa  x,  a  <  x  <  b. 

5.  Express  3  x2  —  2  x  +  5  in  powers  of  x  —  2. 

Ans.   3(x-2)2  +  10O-2)  +  13. 

6.  If  between  two  values  of  x,  namely  x  =  a  and  x  =  b,  a 
function,  f(x),  is  single  valued  and  continuous,  and  has  a  first, 
second,  and  third  derivative,  each  of  which  is  continuous,  then 

Suggestion.     Let 

/(&)-/(«)  -  (b  -  «)/'(«)  -  ^T^/"(«) 

7T— V =  K>  and 

l§ 
F(x)  =  /•(*)  -/(a)  ~(x-  o)/'(a)  -  (-^!/"(«)  -  ^jjp*  K- 

Using  the  results  of  Exercise  6,  express : 

7.  2$  —  3  a2  +  4  #  —  5  in  powers  of  x -f- 1. 

^bis.   2  (x  + 1)3  -  9  (a  + 1)2  + 16  (x  + 1)  - 14. 

8.  3  xz  —  2  x2  —  4  x  +  2  in  powers  of  a;  —  1. 

^ns.    3(x-l)3  +  7(a-l)2  +  (>-l)-l. 


TAYLOR'S   THEOREM  103 

88.  If  between  two  values  of  x,  namely  x  =  a  and  x  =  b,  a 
function,  f(x),  is  single  valued  and  continuous,  and  has  ?i  + 1 
derivatives  all  of  which  are  continuous,  then 

f(p)  =/(«)  +  (b  -  a)/'(a)+fi^r(«)  +  ®^f'"(fi)  +  - 

+  (5_a)»  +  (^-^^(n+i)^   a  <  a^  <  b. 

\n    ■  1 7i  +  l 

This  is  known  as  Taylor's  Theorem.  It  is  readily  seen  to  be 
but  a  more  general  form  of  the  Law  of  the  Mean,  and  the  rjroof 
is  exactly  similar  to  that  given  for  that  theorem  and  for  the 
one  of  Art.  87. 

Proof.     Let 

m^m.-{b-a)f\a)-^^f\a) 

-  (6~a)8/"'(«) (b~^nfnXa) 


|n  +  l 

Since  a  and  b  are  constants,  iTis  a  constant. 
Clear  of  fractions  and  transpose. 


n 

=K  (1) 


_  (b-af    {n)(,  _  (6  -  a)y  K=  ^  (2) 

\n      J     K  J         | Ti  +  1  v  J 


Prom  f(x)  form  the  function 
F(x)  =f(x)  -f(a)  -(x-  a)f'(a)  -  &=^f"(a)  -  &^/' >) 

_  . . .  _  (x  ~  aT  f(nXci)  —  (^  ~  a)n+1  x  CS\ 

\n      J    W        \n  +  l         '  {) 

where  iThas  the  value  given  it  in  (1). 


104       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

F{a)  =/(«)  -/(«)-<>  -  a)f(a)  -  ^)!/»  -  fc^)><(a) 

_  ...  _  \a~  a)n  f(n)(a)  —  (a  ~  a)       j£=  q^ 

\n  \n-\-l 

F(b)  =f(b)  -/(a)  -  (6  -  a)f'(a)  -  (±=f£f»(a)  -  £=22/«'(a) 

(5-a)w   w,     _(t>-a)n+1K=  0 because of (2). 


Differentiate  (3). 
.-.  F'(x)  =f'(x)  -f'(a)  -  (x  -  «)/"(«)  -  fc=-^!/'''(<*)  _ 


2 


-  (a;~a^  -  /™(a)  -  (g7a)n  iT.  (4) 

Since  .F(a)  =  F(b)  =  0,   .-.  F'(^)  =  0,  where  a<\<  b,  by 
Rolle's  Theorem. 

Also,  *"(a)  =/'(a)  -/'(a)-(a-a)/>)-^^V>) 

_  ...      (a  ~  a)w    f(n)(a)  —  (a  ~  a)w  /^_  q 
|w  —  1  [w 

Differentiate  (4). 

.-.  F"(x)  =f"(x)  -f"(a)  -  (a?  -  a)f"'(x) 

_  {X  —  g)w~     f(n)(a\  _  0*?  —  «)W~   j£ 

n-2    ^     W  Ti-1 


Since  .F'(a)  =  i^)  =  0,  .-.  F"(h2)  —  0,  where  a<h2<  h1}  by 
Kolle's  Theorem.  Also,  F"(a)  =  0.  Therefore,  by  Bolle's 
Theorem,  i^"'(/i3)=0,  where  a<li^<h2.  And  thus  we  can  go  on. 
After  n  +  1  differentiations  there  results  F(n+l)(x)=f(n+1\x)  —  K, 


TAYLOR'S   THEOREM  105 

which  is  zero  for  some  value  hn  such  that  a  <  hn  <  hn_1}  where 
/>„_!  is  the  value  of  x  for  which  Fw(x)  =  0. 

Since  a  <  hn <  Jin-\ <  —  <h<  h <h<b,   •'•  ct<  K <  &■ 
...  K=fn+1)(K))   a<hn<b. 

Substitute  in  (1). 
.-.  /(6)-/(a)-(6-a)/'(a)-^)!/"(a)      ■ 

_  (b-affltl(cC) (&-a)"/W(g) 


ra+1 


(&-a) 

[n  +  1 

=  /(n+1W,   a<hn<b. 
Call  /in  =  a?!.     Clear  of  fractions  and  transpose. 

^/(&)=/(a)  +  (6-a)^ 

+  (6 - <*)*jwra\  +  (& ~ a)TC+/("+1)(a?1)?   a<^<6.       (5) 

[w  |  yi  +  1 

89.  In  Equation  (5)  of  the  preceding  article,  we  have  an  ex- 
pression which  holds  true  for  all  values  of  x  from  a  up  to  a 
value  b  at  which  f(x)  or  one  of  its  derivatives  ceases  to  be 
finite  or  continuous.  If  then  we  substitute  x  for  b,  we  have  an 
expression  which  holds  true  from  a  up  to  any  value  x  at  which 
f(x)  or  one  of  its  derivatives  ceases  to  be  finite  or  continuous. 
Substitute  x  for  b. 

+  (*7a)"/""(«)  +  (X~C%+1fn+l\^),  a<x,<x. 
\n  n-\- 1 


This  is  another  form  of  Taylor's  Theorem. 


106       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 
90.    Definitions.     If  f(x)  is  expressed  as 
f(a)  +  (x-  o)/'(o)  +  fej^y»(a)  +fiLj|S2/"'(a)+  ... 

+  fiL=«)!/(»)(a)  +  (a;~a)jl+V(w+1)(^)?  a < ^ < a?, 
|tt  |n  +  l 

it  is  said  to  be  developed  into  a  power  series  in  x  —  a  with  a 

remainder. 

(a;  -  a)n+1 


The  remainder  is 


lw  +  1 


-/(w+1)(Xi)>  a<a7!<a;. 


91.    If  a  is  chosen  zero,  Taylor's  Theorem  becomes 
fix)  =/(0)  +  x/'(0)  + 1/"(0)  + 1/'"(0)  +  - 


71  +  1 


ft 


fn+1\x,),   (XaKx, 


which  is  a  power  series  in  x  with  a  remainder. 

This  special  form  of  Taylor's  Theorem  is  known  as  Maclauren's 
or  Stirling's  Theorem. 

92.    Example  1.     Develop  cos  x  into  a  power  series  in  x  —  a 
consisting  of  five  terms  and  the  remainder. 


f(x)     =  cos  x, 
f'(x)    =  —  since, 
f"(x)  =  —  cos  X, 
f'"(x)  =  sin  x, 
f^ix)  =  cos  x, 
fY(x)  =  —  sin  x, 

.-.  cos  cc  =  cos  a  —  (x  —  a)  sin  a 


.-.  /(a)    =  cos  a. 
.-.  /'(a)   =—  sin  a. 
.•„  /"(a)  =  —  cos  a. 
.*.  /"'(a)  =  sin  a. 
•'•  /Iv(a)—  cos  a- 
•'•  fY(a)  =  ~  s^n  a- 


(x  —  a)4 

IT 


cos  a. 


(a?  — 

*)*< 

12 

(flj  — 

a)5 

cos  a 


(a?  —  «)' 


sin  a 


sin  .%,   a  <  a?!  <  a?. 


TAYLORS   THEOREM 


107 


Example  2.  Develop  sin  x  into  a  power  series  in  x  consist- 
ing of  n  terms  and  the  remainder. 

fix)    Essina,      .-.  /(0)    =0. 

/'(»)   =  cosa,      .-.  /'(0)  =1. 

/"(a;)  =  -sinaj,  .-.  /"(0)  =0. 

/'"(a?) s-oosas,  .'.  /"'(0)  =  -l. 

/^(ae)  =  sin  oj,      .-.  /IV(0)  =  0.  • 

The  law  according  to  which  the  derivatives  proceed  is  that 
all  derivatives  of  even  order  are  sin  x  with  the  +  or  —  sign 
attached,  and  all  of  odd  order  are  cos  x  with  the  +  or  —  sign 
attached.  Also,  that  the  sines  are  alternately  +  and  — ,  and 
also  the  cosines.  Since  each  alternate  term  is  zero,  we  must 
take  2  n  derivatives  in  order  to  get  a  series  of  n  terms  and  the 
remainder. 

.-.  fV*-V(x)  =  (-  l)w"1cos  x,  .-.  /»-«(*)  =  (- 1)"-1  •  1. 
fi2n\x)  =  (-  l)w  sin  x,       .-.  f{2n\x^)  =  (-  l)n  sin  x^. 

/yi3         /y.5  n,2n-l  /v,2n 


x- 


n 


1  \Zn 


0  <  x1  <  x. 

Example  3.     Develop  log  (1  +  a?)  into  a  power  series  in  x 
consisting  of  n  terms  and  the  remainder. 

fix)    =  log  (1 +  *),.-.  /(0)    =0. 
1 


/'(*) 


l  +  x 


/"(*)   = 


(l+»)s 


fin(x)-=J^ 


rwi 


(i+z)3 

1  -2»3 

(1+*)4' 


/'(0)  =1. 
/"(0)=-l. 

/'"(0)=1.2. 

/IV(0)  =  -1.2-3. 


108       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Since  f(x)  =  0,  it  is  necessary  to  take  n  +  1  derivatives  in 
order  to  get  a  series  of  n  terms  and  the  remainder. 

/»(„) ^(-l^-iigi.,  ...  /w(0)  =  (-!)-■  |„ -1. 

/*"(«) = (- 1)"  „  ,M,  ■■■  f("+1K*i)  =  (- 1)" 


^yii         /yi3         ^v>4  ^n 

.-.  log(l  +  a!)=a!-|  +  |-f  +•••  +  (- rr1^ 

l       6       4  w 

+  (-  ly-g-r  •  7=  _/  a,  o  <  x.  <  x. 

?i  +  1     (1  +  x±)n+x 


EXERCISES 

1 .  Express  xi  —  3x3  +  2x2  —  x  +  2  in  powers  of  #  +  2. 

-4ns.  (a?  +  2)4  -  11  (a?  +  2)3  +  44 (a?  +  2)2  _  77  (cc  +  2)  +  52. 

2.  Express  a^-f-Sa;4  —  4a?3  +  2a;2  —  ic  +  2  in  powers  of  x— 1. 
Jim.  (a3-l)5+8(aj-l)4+18(ar-l)3+18(a3-l)2+8(x-l)  +  3. 

3.  Reduce  the  roots  of  the  equation  x5—2x3-\-3x2—x-\-l  =  0 
by  2,  first  by  Homer's  method,  second  by  Taylor's  Theorem. 

Arts.  The  equation  with  reduced  roots  is  x5  +  10  x*  +  38  a?3 
+  71  a;2  +  67  x  +  27  =  0. 

4.  Develop  cos  a;  into  a  power  series  in  a;  consisting  of  n 
terms  and  the  remainder. 

Ans.  cosa?  =  l-^  +  ^r-^+--  +  (-l)n-1 


[2     [4     [6  v       '      [2  71-2 

+  (—  l)nr^ 7 sin  ^  0  <  ^  <  x- 

5.    Develop  sin  a;  into  a  power  series  in  x  —  a  consisting  of 
five  terms  and  the  remainder. 

//£ (£\%    .  (x  —  of)3 

Ans.  sin  a?  =  sin  a  +  (a?  —  a)  cos  a  —      ,0      sin  a  —  * — — -1-  cos  a 
+  (a^-^sina  +  (av-a)_cos^    a<^1<a;. 


TAYLOR'S   THEOREM 


109 


6.    Develop  ex  into  a  power  series  in  x  consisting  of  n  terms 
and  the  remainder. 


a;2  ,  x3  .  .     x71'1     .  x 


Ans.  ex  =  1 +  £+  —  +  — H h 


n  —  1      [n 


|2'18 

7.    Find    the    first    three    terms   of    the    development   of 
log10  cos  x  as  a  power  series  in  x. 

Ans.  The  first  three  terms  of  the  development  are 


-M 


pjl  ry,4  /y»D 

2      12      45 


where  ilf  =  log10  e  =  0.434294  .... 


8.    Develop  ex  into  a  power  series  in  x  +  a  consisting  of 
h,  terms  and  the  remainder. 


Ajis.  ex  =  e 


■1+(.+0)+^+&±ia!+...+fe±^ 


,  (a; +  a)M  x" 
-f-  ^ — ! — ^-  exi 


j     —  (X  <C.  3?i  <C.  37. 


9.    Develop  e3x  into  a  power  series  in  x  consisting  of  n  terms 
and  the  remainder. 


o2™2         03,r3 


Qrc— 1/v.w— 1  SnXn 


[2        [3 


ft  —  1 


[w 


e3a:i 


10.    Find  the  value  of  — ^z-  when  x  =  1. 

a?  —  1 


0  <  ajx  <  a?. 


Suggestion.     Develop  log10a;  into  a  power  series  in  a;— 1  con- 
sisting of  one  term  and  the  remainder.  Ans.  0.434  .... 


11.    Find  the  value  of      xsmx   .  when  x  =  0. 


12.    Find  the  value  of 


x  —  2  sin  a; 

e»  -I-  e-«  _  2 
a;  sin  cc 


en  #  =  0. 


13.  Find  the  value  of    °g  CQS        when  a?  =  0. 

log  cos  a; 

14.  Find  the  value  of  (cos  2  x)x2  when  x  =  0. 
Suggestion.     Let  %  ==  (cos  2  x)x2.      .-.  logw 


-4ws.  0. 
Ans.  1. 
^Ins.  4. 


log  (cos  2%) 
a?2 


110       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Develop  log  (cos  2  x)  into  a  power  series  in  x  consisting  of 

two  terms  and  the  remainder.  1 

Ans.  -=• 

15.  Find  the  valne  of  (a2  —  or)  tan  -p—  when  x  =  a. 

2a  .       4a8 

irx  -4%S. -. 

16.  Find  the  value  of  (3 )         when  x  =  2a.  i. 

93.   In  the  development 
f(x)  =/(a)  +  (x-  a)f\a)  +  &^1>  (a)  +  ...  +  ^=^/^(a) 


ft-h  1 


+ 

the  sum  of  the  first  n  terms  and  the  remainder  is  equal  to  f(x) 
for  all  values  of  x  for  which  f(x)  and  its  first  n  + 1  derivatives 
are  continuous.     Let  n  increase  without  limit.     If 


limit 

n=cc 


(x-a^f{n+1 )( 
n  +  1 


=  0, 


the  limit  of  the  sum  of  the  first  n  terms  is  f(x).  That  is,  the 
series  becomes  an  infinite  series  which  is  convergent  and  con- 
verges to  the  value  f(x).  Then,  for  those  values  of  x  for  which 
the  remainder  approaches  the  limit  zero,  the  function  is  equal 
to  the  infinite  series. 

From  this  infinite  series  the  value  of  the  function  for  par- 
ticular values  of  x  can  be  calculated. 

For  example,  the  development  of  sin  a?  into  a  power  series 
in  x  is 

™3  rJ>  ml  nS.n-\  0£n 


X 

Since  sin  x.  must  be  less  in  absolute  value  than  1,  — —  must 

\2  n 


be  greater  in  absolute  value  than  — —  sin  xv     Now 


x2n  „:„  „       at limit 

n—oo 


2n 


V 


TAYLOR'S   THEOREM 

limit 


is  evidently  zero  for  all  values  of  x.    Therefore 


n=co 


x2n 

sin  a;, 

\2n 


is  zero  for  all  values  of  x.     Therefore,  for  all  values  of  x,  sin  x 
is  equal  to  the  infinite  series 

a»0  /y»0  ^1/ 

To  calculate  the  value  of  sin  x  where  x  has  some  particular 
value,  all  that  is  necessary  is  to  substitute  that  value  in  the 
infinite  series  and  calculate  enough  terms  to  get  the  result  to 
the  required  number  of  decimal  places. 

EXERCISES 

1.  Show  that  cos  x  is  equal  to  the  infinite  series 

/yd  rW*  /yib 

Ill   " 

for  all  values  of  x. 

2.  Show  that  log  (1  +  x)  is  equal  to  the  infinite  series 


if  I x |<  1. 


fp£i  /yvd  /yt^ 

X-2+3-I  + 


CHAPTER   X 
BINOMIAL  THEOKEM 

94.   Develop  (1  +  x)m  into  a  power  series  in  x  consisting  of 
n  +  1  terms  and  the  remainder. 

f(x)  =  (l+x)-,  .'.     /(0)=1. 

f\x)  =  m{l+x)m~\  .'.    /'(0)=m. 

f"(x)  =  m(m-T)(l+x)m-2,       .:  /"(0)=m(m-l). 
f'"(x)  =  m(m-l)(m-2)(l  +  x)m-3, 

.-.  /'"(0)=m(m-l)(m-2). 


f<-n\x)  =  m(m— l)(ra— 2)  •  ••  (m-w  +  1)  (1+a;)' 

.-.  /«(0)=m(m— l)(m— 2)  ...  (m-?i  +  l). 

/c»+D(a;)  =  m(m— 1)  (m - 2)  •  •  •  (m — n  + 1)  (m - n)  (1  +  x)m~n~\ 

.-.  f(n+1)(x1)  =  m(m  —  l)(?n—2)  •  ••  (m— n  +  l)(m  —  ?i)(l+£1)m-M-1 

/-i   .     \™     -i    .  .  m  (m  —  1)    o  ,  m  (m  —  1)  (m  —  2)    .  . 

.-.  (1  +  x)w  =  1  +  ma?  H ^— z  ar  -\ * -^ L  x6  -f  •  •  • 

l£  £ 

m(m  —  l)(m  —  2)  •••(m  —  n  +  1)    n 

+  m(m  ~ *Xm  -  2)  —  (m  -  w  +  l)(m  -  w) ajw+1  ^  +  ^m_K_it 

|n  +  1 

If  m  is  a  positive  integer,  f(m)(x)  =  [m,  a  constant,  and  hence 
all  subsequent  derivatives  are  zero.      In  this  case  the  series 

112 


BINOMIAL    THEOREM  113 

consists  of  a  finite  number,  m  -f 1,  of  terms,  and  (1  +  x)m  is 
equal  to  the  m  + 1  terms  of  the  series 

.,   ,  .  m  (m  —  1)    9  ,  m  (m  —  1)  (m  —  2)    , 

whose  (n  +  l)th  term  is 

m  (m  —  1.)  (m  —  2)  •  •  •  (m  —  n  + 1) 


|n 


or. 


If  m  is  not  a  positive  integer,  the  series  does  not  consist 
of  a  finite  number  of  terms,  but  becomes  an  infinite  series  as  n 
increases  without  limit.  The  infinite  series  is  equal  to  the 
function  (1  +  x)m  for  those  values  of  x}  and  only  those,  for 
which  the  remainder 

m  (m  -  1)  (m  -  2)  ■  ■  ■  (m  -  n  + 1)  (m  -  n)  , 

approaches  zero  as  a  limit  as  n  increases  without  limit. 

It  can  be  proved  without  much  difficulty,  although  the  work 
is  too  long  and  complicated  to  be  given  here,  that  this  re- 
mainder approaches  the  limit  zero  for  all  values  of  x  less  than 
1  and  greater  than  —  1.  We  shall  assume  the  truth  of  this 
statement  and  say  that,  for  those  values  of  x  less  than  1  and 
greater  than  —  1,  (1  +  x)m  is  equal  to  the  infinite  series 

^   ,  .  m(m  —  l)    o  .  m(m  —  l)(m  —  2)    <,  . 

[f  15. 

whose  (n  +  l)th  term  is 

m(m  — l)(m  —  2)  •••  (m  —  n  -f-1) 


xn. 


\n 


95.    The  above  is  but  a  special  case  of  the  development  of 
(a  -f  h)m,  where  a  and  h  are  any  numbers. 


114       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Suppose  that  it  is  required  to  develop  (a  +  h)m.     If  a  is 
greater  than  h,  divide  and  multiply  (a+h)m  by  am.    .-.  (a+h)m  = 

1  _|_  '1  \  .      Since   a  is   greater   than   h,  -  is  less  than   1. 
a)  a 

Let  -  =  x.     .-.  x  is  less  than  1.     Therefore  the  development 
a 

in  Art.  94  holds.     That  is  : 

A    ,  h\m     A  h  ,  m(m  —  l)h2  .  m(m  —  T)(m —  2)7i3  , 

\        aj  a  \2_        a2  [3  a3 

...  (a  +  7*)m=  am+  mcT"^  +  m^  ~  *) a™-27i2 

+  m(m-l)(m-2)am_3/,3+...?  (1) 

n 

the  (n  +  l)th  term  being 

m(m  —  1)  (m  —  2)--  ■  •  (m  —  ?i  + 1)  am-n^n 

If  a  is  less  than  A,  (a  +  7*)m  may  be  written  as  (h  +  a)m. 
Divide  and  multiply  (fc  +  a)m  by  7r. 


...  (ft  +  a)m  =  hmfl  +  £Y 


=  fc-  +  m^"^  +  m(^-1)fr»-aa2 

+  m(m-l)(m-2)fem_8a3+...>    (2) 

I?, 
the  (n  +  l)th  term  being 

m(m  —  l)(m  —  2)  •••  (m  —  n  +  1) /^-n^^ 
[» 

Either  of  the  developments  (1)  or  (2)  is  called  the  Binomial 
Theorem.  It  is  seen  to  be  but  a  special  case  of  Taylor's 
Theorem. 


BINOMIAL    THEOREM  115 

EXERCISES 

1.   Find  the  first  n  terms  of  the  development  of  tan-1  a;  into 
a  power  series  in  x. 

Suggestion.    f(x)  ==  tan-1  x. 

...  f(x)  =  _J_  =  (1  +  x2)-1  =  1  -  x2  +  x*  -  xQ  +  — ,  if  *<1, 

and  f"(x)=—  2x  +  ±xs  —  6x5 -\ ,  if  x<l,  etc. 

Ans.  The  first  n  terms  of  the  development  are  : 

/y»o  /\iO  ryti  /y*Z7l—l. 


a;  — — +  —  —  —  H h(— 1) 


3       5       7  v       J      2n-l 

2.  Find  the  first  four  terms  of  the  development  of  sin-1  a? 
into  a  power  series  in  x. 

Ans.  The  first  four  terms  are  : 

,1     ^,1.3^,1.3.53? 

y\  _| , . 

2     3      2.462-4-67 

3.  Find  the  first  five  terms  of  the  development  of  cos-1  a? 
into  a  power  series  in  x. 

Ans.  The  first  five  terms  are : 

7T  1     x3     1.3a5     1  •  3  •  5  x7 

rQ . . 

2  2     3      2-4  5      2-4-67 

4.  From  the  result  of  Exercise  1,  calculate  w  to  two  decimal 
places. 


CHAPTER  XI 
DIFFERENTIALS 

96.  Up  to  this  point  —  in  the  equation  y  =f(x)  was  treated 

as  a  symbol  and  not  as  a  fraction.  It  is,  however,  sometimes 
convenient  to  treat  it  as  a  fraction,  but  in  order  to  do  so, 
definitions  must  first  be  made  for  dx  and  dy. 

Definitions.  In  a  function  of  one  variable,  the  differential  of 
the  variable  is  defined  to  be  the  same  as  the  increment  of  the 
variable. 

In  a  function  of  one  variable,  the  differential  of  the  function 
is  defined  to  be  the  derivative  of  the  function  with  respect  to 
the  variable  multiplied  by  the  differential  or  increment  of  the 
variable. 

Thus,  in  the  equation  y  =/(#),  the  differential  of  x  is  defined 
to  be  the  same  as  Ax  and  the  differential  of  y  to  be  the  deriva- 
tive of  y  with  respect  to  x,  multiplied  by  either  the  differential 
of  x  or  Asc. 

Differentials  are  denoted  by  writing  d  before  the  variable  or 
function.  Thus,  in  the  equation  y  =/(#),  the  differential  of  x 
is  denoted  by  dx  and  the  differential  of  y  by  dy  or  df(x). 

97.  In  the  equation  y  =f(x), 

dy  =  derivative  x  dx,  by  definition. 

.-.  -M.  =  derivative. 
dx 

Therefore  the  derivative  may  be  looked  upon  as  a  fraction : 
—  the  ratio  of  two  differentials,  as  well  as  a  symbol: — the 
limit  of  the  ratio  of  two  increments. 

116 


DIFFERENTIALS 


117 


The  expression  —  is  the  ratio  of  two  differentials.     Hence 
dx 

the  name  Differential  Calcnlus. 


98.  We  saw  that  the  derivative  is  geometrically  the  slope  of 
the  tangent  line  to  the  curve  at  a  point  (x,  y)  on  the  curve. 
Let  us  see  what  dx  and  dy  are  geometrically. 

Let  P,  (x,  ?/),  be  a  point  on  the  curve.  (Fig.  30.)  Let  PA 
be  the  tangent  line  at  P.  Give  x  the  increment  Ase.  Draw 
the  ordinate  FB  and  pro- 
duce it  to  meet  the  tangent 
line  at  A.  From  P  draw 
PC  parallel  to  the  a>axis  to 
meet  FB  in  C. 

By  definition,  dx  is  the 
same  as  Ax.  Therefore  dx 
is  geometrically  any  arbi- 
trarily chosen  length  DF. 
By  definition,  dy  is  the  de- 
rivative of  y  with  respect  to  x  multiplied  by  dx.  Now  the 
derivative  of  y  with  respect  to  x  is  tan  CPA.  Therefore 
dy = dx  tan  CPA.  Al so,  CA = dx  tan  CPA.  Therefore  dy  =  CA. 
Therefore  dy  is  the  distance,  perpendicular  to  the  cc-axis,  from 
C  to  the  tangent  line. 

It  will  be  remembered  that  A?/  is  the  distance  perpendicular 
to  the  cc-axis  from  C  to  the  curve. 


Y 

^r 

/ 

A 

b" 

c 

/ 

/       0 

[ 

Fj 

3                { 

EG.   30. 

7            X 

99.    Our  formulas  for  finding  derivatives  can  be  readily  con- 
verted into  formulas  in  differentials. 


Example  1.   If 


y  =  u  4-  v, 

dy  _du     dv 

LLtJU        (J/JU         LLJj 


Since  the  derivatives  are  fractions,  we  may  cancel  dx. 

.-.  dy  =  du  -\-dv. 


118       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 
Example  2.     If  y  =  uv, 


dy        dv        du 

\ajJU              \X/Jb              LttX/ 

Cancel  dx. 

.\  dy  =  udv  +  vdu. 

Example  3. 

If 

u 

du        dv 
dy        dx        dx 
dx             v2 

Cancel  dx. 

vdu  —  u  dv 

■•■  */=     v*    • 

EXERCISES 

1.  In  the  equation  y  =  x3,  calculate  Ay  and  dy  when  x  =  2 
and  Ax  =  3.     Represent  geometrically. 

Arts.   Ay  =  117;  dy  =  36. 

2.  In  the   equation  y  =  sinx,  calculate   Ay  and   dy  when 

x  =  - ,  Ax  —  -  •    Eepresent  geometrically. 

Arts.   Ay  =  0.2777;  cfy  =  0.278. 

In  each  of  the  following  equations,  find  dy  for  any  values 
of  x  and  dx. 

r  2#2    ] 

3.  2/ =  a?  log (cc2  + 1).       J.ns.    cfa/  =  Hog  (x?  + 1)  +  — — -  j-  cfo. 

I  ar  + 1  j 

4.  ?/  =  log  sec2  x.  Ans.   dy  =  2  tan  #  cto. 

5 .  y  =  sin2  a;  cos2  #.  u4ws.   dy  —  \  sin  4  #  d#. 

6.  y  =  tan-1  sinh  x.  Ans.   dy  =  sech  x  dx. 
1.   y  =  i  tan3  x  +  tan  x.  Ans.   dy  =  sec4  x  dx. 

8.    y  =  tan-1  log  (a;  + 1).  -4ws.   7 — ,  HNr.    ,  (, — 7 — ,  ,uaV 

*  oV         ;  (a;  +  l)[l-fllog(aj  +  l)j2] 


CHAPTER   XII 

SLOPE  OP  THE  TANGENT  LINE  IN  POLAK   COORDINATES. 

SUBTANGENT.    SUBNOKMAL.    ASYMPTOTES 

100.  We  saw  in  Chapter  III  that  in  the  curve  whose  equa- 
tion is  y  =  f(x)  in  rectangular  coordinates,  where  f(x)  is  single 
valued  and  continuous,  the  slope  of  the  tangent  line  to  the 
curve  at  any  point  (x0)  y0)  on  the  curve  is 


tan  t  =  -^ 
dx 


i 

x=xn 


or,  as  we  shall  write  it, 

tan  r0  =  — 
dx 

To  find  the  slope  in  polar  coordinates  of  the  tangent  line  to 

the  curve  at  a  point  on  the  curve,  we  may  transform  -f-         to 
polar  coordinates  by  means  of  the  formulas  x=x° 

x  =  r  cos  6,  y  =  r  sin  0. 

Since  x  =  r  cos  0,        .'.  dx  =  cos  0  dr  —  r  sin  0  dO. 

Since  y  —  r  sin  6,        .*.  dy  —  sin  6  dr  +  r  cos  6  d6. 

dy  _  sin  0  dr  +  r  cos  6  dO 
dx     cos  Odr  —  r  sin  $d$ 

sin  0 [-  r  cos  0 

cos  0 r  sin  0 

dO 

119 


120        DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


Let  (r0,  0O)  be  the  polar  coordinates  of  the  point  (x0,  yQ). 

Therefore 

•     A  dr 

sin  0O— 

dy 


tan  rn  = 


dx 


dO 


+  r0  cos  0O 


cos  0O 


dr\ 
del 


—  ?*n  sin  On 


is  the  slope  in  polar  coordinates  of  the  tangent  line  to  the  curve 
at  the  point  (r0j  0O). 

If  the  equation  y=f(x)  be  transformed  to  polar  coordinates, 
it  can  be  written  in  the  form  r  =  <j>(0),  or  as  we  shall  write  it, 
r=f(0),  although  f(0)  is  not  the  same  function  of  6  that/(#)  is 
of  x. 

The  slope  of  the  tangent  line  to  the  curve  r  =f(0),  where 
f(6)  is  single  valued  and  continuous,  at  the  point  (r0,  0O),  is 
therefore 


sin  0. 


tan  t0 


dr 

]d0 


+  r0  cos  0O 


cos  6, 


dr 
*d$ 


—  r0  sin  0O 


101.  As  an  illustration  of  the  method  of  finding  the  slope  of 
the  tangent  line  to  the  curve  whose  equation  is  expressed  in 
polar  coordinates,  at  a  particular  point  on  the  curve,  consider 
the  following  example. 

Example.  Find  the  slope  of  the  tangent  line  to  the  curve 
r  =  a  sin  2  0  at  the  point  where  0  =  -  • 


By  the  formula  of  Art.  100, 


O 


135° 


0 


Fig.  31. 


^     A     tan  r0  = 


0  •  sin  -  +  a  cos  - 
4  4 


0  •  cos 


,.  T0  =  135°, 


7T 

a  sm- 
4 


SUBTANGENT.      SUBNORMAL 


121 


102.  The  expression  for  tan  r0  in  polar  coordinates  is  some- 
what complicated.  A  simpler  expression  can  be  found  for 
tan  {f/Q,  where  i//0  is  the  angle  be- 
tween the  radius  vector  to  a  point 
and  the  tangent  line  to  the  curve 
at  the  point.  Let  us  measure  k//0 
from  the  positive  direction  of  the 
radius  vector  to  the  tangent  line, 
in  anti-clockwise  rotation.  Then 
i(/0  is  such  that  tan  ^0=tan(r0—  0O) 
(see  Fig.  32). 

tan  r0  —  tan  0O 


tan  if/0  = 


1  -f-  tan  t0  tan  0O 


sin  0, 


dr 
]d0 


+  n>  cos  0O 


Fig.  32. 


sin  0O 


cos  0O — 
dO 


-,osin0o     C0S^ 


sin0o 


dr 
dO 


+  rn  cos  0( 


o  ^°  ^o 


a  dr 

cos  0O — 

dO 


sin  0n 


rosin0o  oos^ 


e=9n 


dr 
dO 


103.    Subtangent,    Subnormal,    Rectangular    Coordinates.     Let 

V  =/(?)  he  an  equation  in  which  f(x)  is  single  valued  and  con- 
tinuous. Let  PQ  (Fig.  33)  with 
coordinates  (x0,  y0)  be  any  point 
on  the  curve  y=f(x).  Denote 
the  foot  of  the  ordinate  of  P0  by 
M,  and  the  points  where  the  tan- 
gent and  normal  lines  to  the 
curve  at  P0  meet  the  #-axis  by  T 
and  N  respectively. 


122       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


Definitions.     The  lines  TM  and  MN  are  called  the  subtangent 
and  subnormal  respectively  of  the  curve  at  the  point  P0. 


Since 


,  MP, 

tanTo=™> 


TM= 


MP, 

tanr0 


j/o_ 
dy 
dx 


The  subtangent  of  the  curve  at  a  point  (x„  y,)  on  the  curve 
is  therefore  ot 

dy 
dx 

Denote  the  angle  which  the  normal  line  at  P,  makes  with 
the  ;E-axis  by  v,. 


Since 


MP 

tanvn  =  ^-°,     .'.  MN=- 


NM 


MP, 

tan  v. 


Now  tan  v,  =  tan  [  ^  +  t, 


tan  v,  =  —  cot  t0  =  — — 
dy 


.-.  MN-- 


Vo 


dy 
dx 


dx 


The  subnormal  of  the  curve  at  a  point  (x„  y0)  on  the  curve  is 

therefore  ^ 


Vo 


dx 


104.  Subtangent,  Subnormal,  Polar 
Coordinates.  Let  r  =  f(6)  be  an 
equation  in  which  f(&)  is  single 
valued  and  continuous.  Let  P0 
(Fig.  34),  with  coordinates  (?*0,  ^0) 
be  a  point  on  the  curve  r=f(B). 
From  the  pole  0  draw  a  line  per- 
pendicular to  OP,  to  meet  the  tan- 
gent and  normal  lines  to  the  curve 
at  P,  in  T  and  N  respectively. 

Definitions.  The  lines  OT  and 
ON  are  called  the  polar   subtan- 


ASYMPTOTES 


123 


gent   and  polar  subnormal   respectively  of  the   curve   at   the 
point  P0. 

2 

Since     —  =  tan  xb0,     .-.  OT=  OPQ  tan  fa  =  ?f       • 
OP0  Y)  dr 

d®  e=e0 

The  polar  subtangent  of  the  curve  at  a  point  (r0,  $0)  on  the 
curve  is  therefore  2 

d<9 


0  =  0n 


Since 


OiV 


=  tan  NP0O,     .%  ON=OP0tanNP0O. 


OP, 

Now  OPQ  =  rQ,  and, 

tan  NPQ0  =  tan /|-  ^0)  =  cot  fa  =  ™Lfth.    ...  OiV": 


dr 

dO 


The  polar  subnormal  of  the  curve  at  a  point  (r0,  0O)  on  the 

curve  is  therefore  , 

dr 

dd 

105.  Let  y  =/(%)  be  an  equation  in  which  f(x)  is  single  valued 
and  has  infinite  but  not  finite  discontinuities. 

If  the  point  of  contact  of  the  curve  y  =f(%)  and  its  tangent 
line  at  any  point  of  an  infinite  branch  of  the  curve  is  allowed 
to  move  so  that  its  distance  from  the  origin  increases  without 
limit,  the  tangent  line  may  or  may  not  approach  a  limiting 
position.  Thus,  as  will  be  seen  in  Art.  107,  the  tangent  line 
drawn  to  a  hyperbola  thus  approaches  a  limiting  position  while 
that  drawn  to  a  parabola  approaches  no  limiting  position. 

106.  Definition.  If  the  tangent  line  to  a  curve  approaches  a 
limiting  position  as  its  point  of  contact  with  the  curve  moves 
along  an  infinite  branch  so  that  its  distance  from  the  origin  in- 
creases without  limit,  this  limiting  position  is  called  an  asymp- 
tote of  the  curve. 


V2i       DIFFERENTIAL   AND  INTEGRAL    CALCULUS 

The  requirements  that  a  curve  lias  an  asymptote  are  two  in 
number,  namely : 

1st.    That  the  curve  has  an  infinite  branch. 

2d.  That  the  tangent  line  at  a  point  on  an  infinite  branch  of 
the  curve  approaches  a  limiting  position  as  its  point  of  contact 
moves  along  the  branch  so  that  its  distance  from  the  origin 
increases  without  limit. 

If  the  intercepts  of  the  tangent  line,  or  one  intercept  and 
the  slope,  approach  limits  as  the  point  of  contact  of  the  tan- 
gent line  and  the  curve  moves  on  an  infinite  branch  so  that  its 
distance  from  the  origin  increases  without  limit,  these  limits 
determine  the  equation  of  the  asymptote.  If  they  do  not 
approach  limits,  this  branch  of  the  curve  has  no  asymptote. 

107.  The  equation  of  the  tangent  line  to  the  curve  y  =/(*) 
at  a  point  (a^,  y0)  on  the  curve  is 

i/-2/o  =  /         (*-«b> 
d.v  e=st 

Therefore  its  intercepts  are  : 

•*'o  —  Va  y        >  intercept  on  the  .r-axis ; 
d.v 

dy 

Vq  —  -ro  y~,         •  intercept  on  the  ?/-axis. 

r-      v- 
Example    1.      Investigate    the    hyperbola    — ,  — V-,  =  1    for 

.     "  er      0- 


asymptotes.  , 

d.v 


b-.i'o 


0      a~"* 

cr/Av      o-.v0-  —  cru0- 


d.v 


X°       b%~  b%        ' 


,  dy\  lr.iv     cru^  —  hW 

and  //„  —  0%  ;  =  yQ  — 


d.v  x=,o     "        a*y0  «7fo 


ASYMPTOTES 


125 


x  II 

Simplify  by  means  of  the  equality  —^  —  ^  =  1. 


fOO  9        9  9  99  799 

Zr.r, 2  —  q-?/0-  _  g2         ^    a-j/p2  —  &-jy 


b2 


As  the  point  of  contact  of  the  tangent  line  moves  along  the 
curve  so  that  its  distance  from  the  origin  increases  without 
limit,  the  coordinates  both  increase  without  limit.  If  (a*0,  y0) 
are  the  coordinates  of  the  moving  point, 


limit 


a- 

x0 


=  0,    and    limit 


b2' 


=  0. 


Therefore  the  hyperbola  has  an  asymptote. 

b2xn 


Since 


dy 
dx 


a-Vo 


=  ± 


bxn 


a  \Xq 


b 


4- 


as 


limit 


dy 
dx 


limit 


a 


\ 


1- 


cv 

X(\ 


b 
=  ±  - 
a 


The  hyperbola  has  therefore  two  asymptotes,  one  with  slope 

+  -  and  the  other  with  slope .     Their  equations  can  be 

a  x  a 

written  readily  from  analytic  geometry.     They  are : 

a      b        ' 


and 


X  V  rx 

-  +  f  =  0. 

a      o 


Example  2.     Investigate  the  parabola  y2  =  2  mx  for  asymp- 
totes. 


dy 
dx 


m 


126       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


y<f 


and 


Now 


■• 

^o  ~~Vo 

dy 

0      m 

—  X0) 

dx 

x  =  x0 

dy 
y°~X°dx 

_         mx0__y0^ 
x=x0               Vo       % 

it 

—  3?o 

= 

-oo,  and    limit 

_2_ 

=  00. 


Therefore  neither  intercept  approaches  a  limit,  and  therefore 
the  parabola  has  no  asymptote. 

108.  If  the  polar  subtangent  OT  (Fig.  34)  approaches  a  limit 
as  the  point  P0  moves  along  an  infinite  branch  so  that  its  dis- 
tance from  the  pole  increases  without  limit,  the  curve  has  an 
asymptote  passing  through  T  in  the  limiting  position  of  OT 
and  parallel  to  OP0. 

Example.    Investigate  the  curve  r=a  sec  2  0  for  asymptotes. 
Since  r  =  a  sec  2  0, 

.-.  —  =  2asec2  0tan2  0. 
dO 

r2      a  sec 20     a  „  . 

•*•  7T=oz — ^  =  oCosec20. 
dr      2  tan  2  6      2 

dO 


As  the  point  of  contact  of  the  tangent  line  moves  along  the 
curve  so  that  its  distance  from   the   pole   increases   without 

7tt 


limit,  0  approaches  either    j,    — -,    — - 


or 


Now 


limit  r 


cosec  2  6 


a 

25 


limit 
4 


-  cosec  2  6 

2 


a 


limit 

0==^ 
4 


-  cosec  2  9 


a 


limit 
4 


a 


cosec  2  6 


ASYMPTOTES 


127 


The  curve  has  therefore  four  asymptotes,  two  intersecting  on 
the   initial   line   at   a   dis- 

ft 

tance from  the  pole  and 

making  angles  of  45°  and 
315°  respectively  with  the 
initial  line,  and  two  inter- 
secting on  the  initial  line 

at  a  distance from 

V2 
the  pole  and  making  angles 

of  135°  and  225°  respec- 
tively with  the  initial  line. 

The  curve  is  as  in  Fig.  35. 

Fig.  35. 


EXERCISES 


1.  Find  the  slope   of  the   cardioid   r  =  a(l  —  cosO)   when 

0  =  -.  Ans.    -1. 

2 

2.  Find  the  slope  of  the  curve  r  =  a  sec2  6  when  r  =  2  a. 

Ans.   3. 

3.  In  the  circle  r  =  a  cos  6,  find  t0  and  \pQ. 

Ans.  If  6  varies  between  0  and  it, 

TO  =  ±!  +  20o,or  _^  +  20o;  ^0=±|  +  ft. 

4.  In  the  logarithmic  spiral  r  =  and,  show  that  if/  is  constant. 

5.  In  the  lemniscate  r2  =  a2  sin  2  6,  find  r0  and  i/-0. 

Ans.   If  0  varies  between  0  and  -, 

r0  =  3  00,  or  —  7T  -f  3  00 ;  i^0  =  2  0O. 

In  each  of  the  three  following  curves,  find  the  subtangent 
and  subnormal  at  any  point  (xQ,  y0)  on  the  curve. 

6.  y2  =  4zax.  Ans.   2x0-,  2a. 


7.    xy  =  c. 


jUns.    ■—  Xq  '.   — 


yj 


128       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

8.  y2=      ^      ■  Ans.   2ax^~x^    3  gay-ay* 

2  a  — x  3  a  —  x0       (2  a  —  x0y2 

9.  In   the    cycloid    x  =  a (0  —  sin  0),    y  =  a(l  —  cos  0),    find 
the  subtangent  and  subnormal  for  a  value  0O  of  0. 

2~  ~  2 


Ans.   2asin3— °sec^;  asin0o. 


10.  In  the  curve  x  =  a  log  cot  -  —  a  cos  0,y  =  a  sin  0,  find  the 
subtangent  for  a  value  0O  of  0-  Ans.    —  a  cos  0O. 

11 .  In  the  lemniscate  i^—a2  sin  2  0,  find  the  polar  subtangent 
for  a  value  0O  of  0.  .4ws.   a  tan  2  0O  Vsin  2  0O. 

12.  In  the  curve  r  =  asec20,  find  the  polar  subnormal  for  a 
value  0O  of  0.  Ans.   2  a  sec2  0O  tan  0O. 

Investigate  the  following  curves  for  asymptotes. 

13.  y2  —  x'  —  3x  —  4.  ,4ns.   Asymptotes,  ~    ' 

2x-\-2y  =  3. 

14 .  2Z3  =  ^  —  8  a?2.  Ans.   Asymptote,  x  —  y  =  f . 

15 .  ?/2  = ^4?is.   Asymptote,  x  =  2a. 

2  a  — x 

16.  r  =  «tan0.     As.  Two  asymptotes  perpendicular  to  OA 
and  distant  a  from  0. 

a 

17.  r  =  -  •     Ans.   An  asymptote  parallel  to  OA  at  a  distance 

a  above. 


CHAPTER   XIII 

CURVATURE.  RADIUS  OP  CURVATURE. 
EVOLUTES  AND  INVOLUTES 


109.  Definitions.  The  integral  curvature  of  an  arc  of  a  curve 
whose  equation  gives  a  continuous  function  is  the  angle  between 
the  tangents  at  the  extremities  of  the  arc. 

Thus,  if  P0P  (Fig.  36)  be  the  arc,  and  APQ  and  BP  the  tan- 
gents at  its  extremities,  the  integral 
curvature  is  the  angle  ACB. 

The  mean  curvature  of  an  arc  of 
a  curve  whose  equation  gives  a  con- 
tinuous function  is  the  integral  cur- 
vature of  the  arc  divided  by  the 
arc,  the  integral  curvature  being 
measured  in  units  of  angle  and  the 
arc  in  units  of  length. 


Thus,  in  Fig.  36,  the  mean  curvature  is 


A  ACB 


arc  P0P 

Mean  curvature  is  usually  expressed  in  radians  per  unit  of 
length. 

The  actual  curvature  at  a  point  on  a  curve  whose  equation 
gives  a  continuous  function  is  the  limit  which  the  mean  curva- 
ture of  an  arc  beginning  at  the  point  approaches  as  the  arc  is 
allowed  to  decrease  without  limit. 

Thus,  in  Fig.  36,  the  actual  curvature  of  the  curve  at  the 


point  P0  is        ^J*    . 
r  arc  P0P  =  0 

For    brevity,    actua 

curvature. 

K 


ZACB 
arc  PqP 

curvature    is    spoken    of    merely   as 


129 


130       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


The  curvature  of  an  arc  is  said  to  be  uniform  when  it  is  the 
same  at  all  points  of  the  arc. 

110.    Let  y  =/(#)  be  an  equation  in  which  f(x)  is  single 

valued   and   continuous.      To   find  the  curvature  at  a  point 

(x0,  y0)  on  the  curve  y=f(x). 
Suppose  that  the  curve  is  as  in  Fig.  37.     Let  P0  be  the  point 

whose  coordinates  are  (a?0,  y0). 
Let  x  =  x0  +  Ax.  Let  P  be  the 
point  whose  abscissa  is  x0  +  Ax. 
Denote  the  arc  P0P  by  As  and  the 
chord  P0P  by  Ac. 

Denote  the  angle  which  the  tan- 
gent line  at  PQ  makes  with  the 
#-axis  by  r0.  The  angle  which  the 
tangent  line  at  P  makes  with 
the    o>axis    is    therefore    t0  +  At, 

where  At  is  the  increment  in  the  angle  due  to  x  having  taken 

the  increment  Ace. 


The  curvature  at  P0  =  }m{\ 


"At" 
As 


,  limit 


At" 

As 


,  since  Ax  =  0  as  As  =  0. 


It  will  be  shown  in  Chapter  XXII  that  the  limit  of  the  ratio 
of  a  chord  to  its  arc  as  both  approach  zero  is  1.  Assuming 
this  theorem,  we  have  : 


The  curvature  at  Pn  =  lim?ti 


Aar  =  0 


At 

Ac 


Ac 
As 


.  limit 
A:r=0 


At 

Ac 


limit 
Az  =  0 


Ac 
As 


.  limit 
A#  =  0 


"At" 

Ac 


since 


limit 
Ax  =  0 


"Ac" 
As 


=  1, 


CUBVATUHE 


131 


limit 

Az  =  0 


limit 
Ax  =  0 


_  VAx2  +  A/ J 


At 

Ax 


n-(s: 


l 


by  dividing  and  multiplying  by  Aa, 

a) 


dec 


To  express 


dr 
dx 


*         \dsc  *=*oy 


in  terms  of  a;  and  y. 


tanT  = 


-<M. 


dx 


o    dr      d2y 
.*.  secJ  r  ■ —  =  —J- 


dr 


d2y 
dx2 


d2y 
sec2  rn       1  +  tan2  rn 


d2y 
dx2 


\dx 

Substitute  this  value  in  (1). 
Therefore  the  curvature  at  P0  = 


dx- 


\dx 


J) 


(2) 


Curvature  is  usually  denoted  by  k. 


that  of  5 y 

dxr 


132       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

111.  In  the  expression  (2)  for  curvature,  in  the  preceding 
article,  the  positive  square  root  was  taken.  Then,  since  the 
denominator  is  always  positive,  the  sign  of  k  will  depend  on 

•     Therefore  k  is  positive  if  the  curve  is  con- 

cave  upwards,  and  negative  if  concave  downwards. 

There  is  no  advantage  in  distinguishing  between  positive 
and  negative  curvature,  and  consequently,  when  the  curvature 
is  negative,  we  shall  omit  the  minus  sign  and  consider  it  as 
positive. 

112.  As  an  illustration  of  the  method  of  finding  the  curva- 
ture at  a  given  point  on  a  given  curve,  consider  the  following 
example. 

Example.  Find  the  curvature  at  a  point  (x0,  y0)  on  the 
circle  x2  +  y~  =  <$> 


Since  x2  +  y2  =  a2,  .*.  y  —  ±  Va2  —  x2. 

At  first  consider  the  part  of  the  curve  above  the  a^axis. 


Then  y  =  +  Va2  —  x\ 


•   dy  =  x  and  dy  =  a 

"  dx         Va2-^'  <&*  (a2 -a2)* 


a2 


(a2-x2y  1 

K  = 


f  Xq  }  2 

{         a2-x02) 


,2      if  a 


Or,  as  we  sharll  say,  #c  =  -• 


a 


Next  consider  the  part  of  the  curve  below  the  a-axis. 
Then  y  =  —  Va2  —  x2. 


CURVATURE 


133 


:,  and  —4  = 


dy x_ 

dx~  Va2^'  '     "  d^     (a2  -  x2) 


af 


'.     K  = 


(a2-<P 


1  + 


,.2 


a2  —  x02 


I      a 


113.  Let  r—f{6)  be  an  equation  in  which  f(6)  is  single 
Valued  and  continuous.  To  find  the  curvature  at  a  point 
(?o>  0o)  on  tne  curve  r=f(0). 

Suppose  that  the  curve  is  as  in  Fig.  38. 
Let  P0  be  the  point  whose  coordinates 
are  (r0,  0O).  Let  6  =  00  +  A<9.  Let  P  be 
the  point  at  which  9  =  6Q  +  A#.  De- 
note the  arc  P0^  by  As  and  the  chord 
PqP  by  Ac.  Let  the  angles  which  the 
tangent  lines  at  P0  and  P  make  with 
the  initial  line  be  denoted  by  t0  and 
t0  +  At  respectively. 

Suppose  that  the  equation  r  =  f($)  is  transformed  to  rec- 
tangular coordinates.     The  equations  of  transformation  are 


Fig.  38. 


x  =  r  cos  0, 
y  =  r  sin  6. 


If  P0  is  the  point  whose  abscissa  is  x0)  and  P  the  point  whose 
abscissa  is  x0  +  A#, 


__  limit 
Az  =  0 


"At" 

As 


limit 
Az=0 


_  VAa?2  -f  A?/ 


3 


Now  A0  =  O  as  Aa?  =  0. 


134       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


Therefore 


K  = 


limit 
A0  =  O 


At 


limit 
A0  =  O 


_  VAa;2  +  Ay2_ 

At 

A0 


vrs) 


AA2/Y 


w+^jj 


by  dividing  and  multiplying  by  A#? 


d0 


2+fdy 


Since 


Vfdx  I 
tan  t  =  tan  (0  +  if/)  (see  Art.  102), 


(1) 


.-.  T  =  &7r  +  0  +  ^,  where  ft  is  zero  or  an  integer  positive  or 
negative. 


d0 


d0 


Since 


tan  \p  =  —  (see  Art.  102), 


d0 


.*.  ^  =  tan_1 


dr 
dd 


d0~~ 


e2rV  _  9  c22r 

dOj  ~7  dO2 

\dOJ 


dr 

dO 


r02  +  2 


(dr 

\d$ 


2        (2V 


d02 


0=0n 


0  =  0n 


^s 


CURVATURE 


135 


Also,  since  x  =  r  cos  0,  .'.  ax—  cos  0  dr  —  r  sin 6  dO ;  and  since 
y  =  r  sin  0,  .*.  dy  =  sin  0  dr  +  r  cos  0  dO. 


\fdx\ 


[dO 


=V(< 


d<9 


—  r0  sm 


*«Y+(-*o| 


+  r0  cos  0O 


+ 


dO 


Substitute  the  value  of 


dr 
dO 


0=6, 


in  (1). 


114.  As  was  seen  in  Art.  112,  the  curvature  of  a  circle  is 
the  reciprocal  of  the  radius  of  the  circle.  Then,  since  a  circle 
can  be  described  with  any  radius,  a  circle  can  be  described  to 
have  any  curvature. 

115.  Definitions.  Two  curves  are  said  to  be  tangent  to  each 
other  at  a  point  when  they  coincide  at  the  point  and  have  the 
same  tangent  line  there. 

The  circle  tangent  to  the  curve  and  having  the  same  curva- 
ture as  the  curve  at  a  point,  whose  concavity  has  the  same 
aspect  as  that  of  the  curve,  is  called  the  osculating  circle  of 
the  curve  at  the  point. 

The  radius  of  the  osculating  circle  at  a  point  on  a  curve  is 
called  the  radius  of  curvature  of  the  curve  at  the  point. 


136       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

The  center  of  the  osculating  circle  at  a  point  on  a  curve  is 
called  the  center  of  curvature  of  the  curve  at  the  point. 

From  its  definition,  the  radius  of  curvature  of  a  curve  at  a 
point  is  the  reciprocal  of  the  curvature  at  the  point. 


EVOLUTES 

116.   Definition.     The  locus  of  the  center  of  curvature  of  a 
given  curve  is  called  the  evolute  of  the  curve. 

Let  y  =f(x)  be  an  equation  in  which  f(x)  is  single  valued 

and  continuous.     To  find  the  evolute  of  the  curve  y  =f(x). 

Suppose  that  the  curve  is  concave  upwards  and  rising,  as  in 

Fig.  39.     From  the  investigation  in  this  case,  the  student  can 

readily   investigate   the    other 

cases  for  himself.     It  will  be 

found    that    in   all    cases    the 

results  are  the  same. 

Let     D     with     coordinates 

(x0,  y0)  be  a  point  on  the  curve, 

and  F  with  coordinates  (x',  y') 

the    corresponding    center    of 
Fig.  39.  , 

curvature.     Denote  the  radius 

of  curvature  of  the  curve  at  D  by  p.     From  D  draw  DC  paral- 
lel to  the  ic-axis  to  meet  the  ordinate  y'  in  C. 


Then 
and 

Since 

.*.     COS  Vq  = 


X   —  Xq  —  Oxy, 

y'  =  y0+CF. 


1 
vo  =  t0  +  2>  •'•  tanv0  =  —  -^~ 

dx 


dy 
dx 


^         V  dx 


:,  and  sin  v0  = 


V>+(! 


dy 
dx 


EVOLUTES  AND  INVOLUTES 


137 


Now  CD  —  |  p  cos  v0 1  = 


\dx 


2lf 


<Q/ 
da?2 


v*+fi 


\dx 


dx 


\dx 


dx2 


because 


dy 
dx 


is  positive  since  the  curve  is  rising,  and 


fy 

dx2 


is  positive  since  the  curve  is  concave  upwards 

\dx 


Also, 


dx 


y'=yo  +  \psmVo\ 


d2y 
dx2 


(1) 


1  + 


y  =2/0  + 


dy 

dx 


d2y 


(2) 


If   x0  and   y0  be   eliminated   from   equations  (1),  (2),  and 
2/o  =/(a?0),  the  resulting  equation  is  that  of  the  e volute. 

117.    Example  1.    Find  the  e volute  of  the  parabola  y2=2  mx. 
At  first  take  y  =  +  V2  m  Va?  and  consider  the  part  of  the 
curve  above  the  a>axis. 

dy  _     /ml 
dx       * 


2  v: 


iC 


d2y_   _  1    /m  _1 
cta2~      2^2^ 


138       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


and 


m 


'-a,  -Jm    1  2x» 

*2  Vab       1     m   1 


=  3  x0  +  m, 


2  X  2  ^ 


m 


2/  =2/o-|- 


9  o* 


_1    Imi 

2\  2  ~f 


m2 


a) 


%  — 


a;'  —  m 


,  and  y0  =  —  VmLy 


Substitute  the  values  of  x0  and  y0  in  2/o2  =  2  maj0 


my 


'2  —  _B_f~i 


27 


(a;'  —  m)3  is  the  equation  of  the  evolute. 


Solve  equation  (2)  for  y1.     .*.  y 


I 


2V2 
3V3 


(x'  —  m)2. 


m 


(2) 


In  this  equation  the  minus  sign  must  be  taken,  since,  from 
equation  (1),  y'  is  negative.  The  part  of  the  parabola  above 
the  a3-axis  and  its  evolute  are  therefore  as  in  Fig.  40. 


Fig.  40. 


EVOLUTES  AND   INVOLUTES 


139 


Similarly  we  may  show  that  the  evolute  of  the  part  of  the 

parabola  below  the  #-axis  has  the  equation  y'  =  -\ — —  (x'  —  m)*- 

3V3m 

The  parabola  and  entire  evolute  are  as  in  Fig.  41. 

Example  2.     Find  the  evolute  of  the  cycloid 
x  =  a(6  —  sin 0),   y  =  a(l  —  cos 0). 
dy  _      sin 


dx 

1- 

-cos  0 

d?y  = 
dx2 

d  - 
"dO^ 

(    sin  (9    > 
\Jl  —  cos  Bj 

'  dx 

1 

a  (1  —  cos  ^)5 


Fig.  42. 


.-.    a?'  =  a  (0O  —  sin  0O)  +  2  a  sin  0O  =  a  (Q0  +  sin  00). 

y'  =  a  (1  —  cos  0O)  —  2  a  (1  —  cos  0O)  =  —  a  (1  —  cos  0O). 
Therefore  the  equations  of  the  evolute  are 

x'  =  a(0  +  sin  0), 

y'  =  —  a(l  —  cos  0). 
One  arch  of  the  cycloid,  and  its  evolute,  are  as  in  Fig.  42. 


PROPERTY  OF   THE  EVOLUTE 

118.  The  evolute  of  a  curve  has  the  property  that  the 
extremity  of  a  stretched  string  unwound  from  it  traces  out 
the  original  curve. 

Thus,  in  Fig.  40,  Art.  117,  the  extremity  0  of  a  stretched 
string  OBC  unwound  from  BC  traces  out  the  part  of  the 
parabola  OA. 

On  account  of  the  property  of  the  evolute  just  mentioned 
the  original  curve  is  called  the  involute. 


140       DIFFERENTIAL  AND  INTEGRAL    CALCULUS 

119.    To  prove  the  above  property  of  the  evolute  we  have 
to  establish  two  theorems. 


Theorem  I. 

evolute. 


Every  normal  to  the  involute  is  tangent  to  the 


Theorem  II.  The  length  of  any  arc  of  the  evolute  is  equal 
to  the  difference  between  the  lengths  of  the  radii  of  curvature 
of  the  involute  which  pass  through  the  extremities  of  the  arc 
in  question. 

Proof  of  Theorem  I. 

Let  D  with  coordinates  (x,  y)  be  a  point  on  the  involute, 

and  F  with  coordinates  (x',  y'), 
the  corresponding  point  on  the 
evolute.     (Fig.  43.) 

Let  the  tangent  line  at  F  and 
the  normal  line  at  D  make  the 
angles  r'  and  v  respectively  with 
the  a>axis.  The  theorem  is  proved 
when  it  is  shown  that  tan  r'  =  tan  y. 


Fig.  43. 


V^" 


dx 
dx 


dx' 


From  (1)  and  (2)  of  Art.  116, 


x'  =  x 


2/'  =  2M 


dy 

\dxj 

dx           dry 
dx2 

i          s 

-r 

d2y 
dx2 

E VOLUTES  AND  INVOLUTES 


141 


dy  fd2y\2  d3y 
dy'  dx  \dx2)  dx3 
dx  ~  fd2y\2 

dx2) 


dyV  d3y 
dx)  dx3 


j  ax 

and  —  =  — 


dx' 
dx 


dy' 


dx  \    dx  \dx2J       dx3      \dxj  dx3  j 
dx2 


dx  1    .      _.   .  . 

•'•^=-%'bydlvlsl0n- 

dec  dx 

.-.   tanr'  =  tanv. 

Proof  of  Theorem  II. 

Let  D  with  coordinates  (x,  y)  be  a  point  on  the  involute, 
and  F  with  coordinates  (x',  y')  the  corresponding  point  on  the 
evolute.  Let  x  take  an  increment 
Ace.  Let  E  be  the  point  on  the 
involute  whose  coordinates  are 
(x  +  Ax,  y  +  Ay),  and  G  the  corre- 
sponding point  on  the  evolute. 
Let  (x'-\-Ax',  y'-\-Ay')  denote  the 
coordinates  of  G.  Denote  the 
arc  FG  by  As',  and  the  chord  FG 
by  Ac'.  Fig.  44. 


Now 


limit 


As' 

Ax 

Ac'     As' 

=A5-A?'identically 

"As'"1 

_  limit 

Ac' 

limit 

As'" 

Ax  ^ 

Ax  =  0 

Ax 

Ax  =  0 

Ac'_ 

limit 

"Ac'" 
Aa? 

,  since  ll] 

'                Ax 

nit 
=  0 

"As' 
_Ac~'_ 

_  limit 
Ax  =  0 

~Va 

x'2  +  Ay'T 

Ax 

• 

=  1  (see  Art.  194) 


142       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


ds'  _      fdx'X2      fdy'^2 
dx~*\dx  J       \dx 


3 dy  fd2y\2 _d?y_  fdyX2  d?y 


Vfdy\2  v  dx  \dxrj       dx3      \dxj  dx3 


From  the  formula  of  Art.  110, 

.  (i+(i)T 


dx2) 


P  =  Z  = 


d2y 
dx2 


dPy 

dp         L       (dyX2  "  dx  \dx2J       dx3      \dxj  dx8 


-V*+ffi 


3  ty  f<3W  _^y_(dy_ 

yX2     dx  \dx2J       dx3      \dx 


dx       *         \dxj  (d2y\2 

\M) 
ds'      dp 

U/Jb  LttV 

It   will    be    seen    later   that   since    -—  =  —-.   .-.  s'  =  p  +  c, 

dx       dx  r 

where  c  is  any  constant.     The  student  can  see  now  that  if 
s'  =  p  +  c   be  differentiated  with  respect   to  x,  the  result  is 
ds'  _  dp 
dx      dx 

Let  Si  and  s2'  be  the  lengths  of  arcs  of  the  evolute  measured 
from  some  fixed  point  on  it,  and  p1  and  p2  the  radii  of  curvature, 
which  pass  through  the  extremities  of  these  arcs. 


Then,  since 

s'  =  P  +  c, 

we  have 

*l'  =  Pi  +  c, 

and 

$2    =  P-2  +  C 

-:.-.  V 

—  -V  =  Pi  —  P2 

Now  Si  —  s2'  is  any  length  of  arc,  and  px  —  p2  is  the  differ- 
ence between  the  radii  of  curvature  which  pass  through  the 
extremities  of  the  arc.     The  theorem  is  therefore  proved. 


EVOLUTES  AND  INVOLUTES  143 

EXERCISES 

In  each  of  the  two  following  curves,  find  the  curvature  at  a 
point  (0*0,  y0)  on  the  curve. 

7)% 

1.    if  =  2mx.  Ans.  k  = 


9  ,      9  >  I 


,2        2  f/"V>4 

2.  £.4-2/_  =  l.  ^[ns.  K  = . -. 

In  each  of  the  two  following  curves,  find  the  curvature  at  a 

point  (r0,  0O)  on  the  curve. 

3 

3.  r  =  a(l  —  cos  0).  Ans.  k  = 


2V2 


ai 


06  A                          1              3  00 

4.  r  =  a  sec2  -•  ^4ws.  k  =  — -  cos°  ^- 

2  2  a          2 

In  each  of  the  six  following  curves,  find  the  radius  of  curva- 
ture, p,  at  a  point  (x0,  yQ)  on  the  curve. 

5.  xy  =  c.  Ans.  P  =  ^±1^. 

6.  °?-y-  =  l.  Ans.  p=j6V  +  «VJ*. 
a2     62  a4*4 

-          --  V  2 

7.  ?/  =  ^(ea  +  e  a).  -4ws.  p  =  ^- 


2  2  2 


8.  cc3  +  ?/3=a3.  ^4ws.  p  =  3(a^o2/o)3- 

9.  ex  =  siny.  Ans.  p  =  e~xo. 

3 

10.  a2+?/2=a^  Ans.  p  =    v  otjo;  . 

11.  Find  the  radius  of  curvature,  p,  of  the  cycloid 

x  =  a(0  —  sin  0),  y  =  a(l  —  cos  6)  for  a  value  60  of  0. 

ZD 

Ans.  p  =  4ca  sin  —  ■ 

H  2 

12.  Find  the  radii  of  curvature  of  the  trochoid  x= aO+k  sin  0, 
y  =  a  —  k  cos  0  at  the  points  where  it  is  nearest  to  and  farthest 
from  the  base.  ^ns   (a  ±  ~k)2 


144       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

In  each  of  the  three  following  equations  find  the  radius  of 
curvature,  p,  at  a  point  (r0,  0O)  on  the  curve : 

13.    r*=  a2  cos  2  6.  Arts.  p  = 


3r0 


,  3 


14.    r2  cos  2  (9  =  a2.  ^is.  p  =  \ 


2  r$ 

15.  r(l  -f  cos  0)  =  2  a.  Ans.  p  =  =-1^- 

In  each  of  the  six  following  curves,  find  the  equation  of  the 
evolute  of  the  curve  : 

16.  x2  +  y2  =  a2.  Ans.  x'  =  0,  yf  =  0. 

17.  ?L  +  t  =  l.  Ans.  (ax')%  +  (by')%  =  (a2-b2)i 
a2     b2 

18.  t-t  =  1.  Ans.  (ax')%-(by')%=(a2+V*)%. 

a2     b- 

19.  **  +  yt  =  cA  A*   p^1  +  (^)f  =2. 

20.  ajy=c.  -4ns.  (aj'+y*)*— («'— y')*=2(2c)*. 

21.  a;  =  a  (cos  <£  -f  <£  sin  <£),  .y  =  a  (sin  </>  —  <f>  cos  <£). 

^4ns.  x'  -\-y'  =a2. 

In  each  of  the  following  curves,  find  the  coordinates  of  the 
center  of  curvature  of  the  curve  at  a  point  (x0,  y0)  on  the  curve  : 

22.  xs  =  ay2.  Ans.  x'  =  —  x0  —  — ^-, 

2  a 


y'=±U+£)4 


a 


23.    y  —  -  (ea  +  e  a).  Ans.  x'  =  x0  —  ^ Vy02  —  a2, 

2  a 

2/'  =  2  2/o- 


CHAPTER  XIV 

DIFFERENTIATION  OF  FUNCTIONS  OF  TWO  INDEPENDENT 

VARIABLES 

120.  In  the  preceding  chapters,  the  functions  considered 
were  all  of  one  variable.  The  functions,  however,  with  which 
we  have  to  deal  are  sometimes  functions  of  two  or  more  inde- 
pendent variables.  Thus,  for  example,  the  area  of  a  rectangle 
is  a  function  of  two  independent  variables :  —  the  two  sides ; 
and  the  volume  of  a  rectangular  parallelopiped  is  a  function  of 
three  independent  variables  :  —  the  three  edges.  In  this  chap- 
ter the  differentiation  of  functions  of  two  independent  vari- 
ables will  be  considered. 

121.  In  a  function,  f(x,  y),  of  two  independent  variables,  if 
real  values  be  assigned  to  x  and  y,  and  the  resulting  value  or 
values  of  f(x,  y)  be  real,  the  values  for  x,  y,  and  z  =f(x,  y), 
plotted  according  to  the  conventions  of  analytic  geometry  of 
three  dimensions,  determine  a  point,  or  points,  in  space.  Then 
z  =  f(x,  y)  is  geometrically  represented  by  a  surface,  or  by 
surfaces. 

For  example,  in  the  function  ±  Va2  —  x?  —  y2,  given  by  the 
equation  z  =  ±  Va2  —  x2  —  y2,  if  real  values  be  assigned  to  x 
and  y,  such  that  x2  +  y2  <  a2,  ±  Va2  —  x2  —  y2  has  real  values, 
and  the  values  of  x,  y,  and  ±  V<x2  —  x2  —  y2,  plotted  on  the 
axes  of  x,  y,  and  z  according  to  the  conventions  of  analytic 
geometry  of  three  dimensions  in  rectangular  coordinates,  deter- 
mine two  points  in  space.  Then  z  =  ±  Va2  —  x2  —  y2  is  geo- 
metrically represented  by  two  surfaces.  As  a  matter  of  fact, 
l  145 


f(a  +  h,   b  +  k) 


=/(«,  &)> 


146       DIFFERENTIAL  AND  INTEGRAL    CALCULUS 

in  this  particular  example,  the  two  surfaces  join  together  and 
give  the  entire  surface  of  a  sphere  with  center  at  origin  and 
radius  a, 

122.  Definitions.  A  function,  f(x,  y),  of  the  two  independent 
variables  x  and  y,  is  said  to  be  single  valued  when,  for  values  of 
x  and  y,  the  function  has  just  one  value. 

A  single-valued  function,  f(x,  y),  of  the  two  independent 
variables  x  and  y  is  said  to  be  finite  when  there  is  no  set  of 
values  of  x  and  y  for  which  the  function  does  not  have  a 
definite  value. 

A  single-valued  function,  f(x,  y),  of  the  two  independent 

variables  x  and  y,  is  said  to  be  continuous  for  the  set  of  values  a 

limit 
and  b,  when  f(a,  b)  is  finite,  and  h=0 

Jc=0 

no  matter  how  7i  and  k  approach  zero. 

In  what  follows  we  shall  suppose  that  f(x,  y)  is  single  valued 
and  continuous. 

123.  In  a  function,  f(x,  y),  of  two  independent  variables, 
we  have  to  consider  : 

Case  I.  How  fix,  y)  varies  when  x  varies  and  y  remains 
constant ; 

Case  II.  How  f(x,  y)  varies  when  x  remains  constant  and 
y  varies ; 

Case  III.  How  f(x,  y)  varies  when  both  x  and  y  vary  and 
are  independent  of  each  other. 

124.  It  was  seen  above  that  z  =  f(x,  y)  in  Case  III  is  repre- 
sented geometrically  by  a  surface.  In  Case  I,  where  y  is  con- 
stant, z—f(x,  y)  is  represented  geometrically  by  a  plane  curve 
formed  by  the  intersection  of  the  plane  y  =  the  constant,  with 
this  surface.  In  Case  II,  where  x  is  constant,  z  =  f(x,  y)  is 
represented  geometrically  by  the  plane  curve  formed  by  the 
intersection  of  the  plane  x  =  the  constant,  with  this  surface. 


TWO  INDEPENDENT   VAUIABLES 


147 


125.  In  either  Case  I  or  Case  II  the  function  f(x,y)  is 
exactly  like  those  considered  in  the  preceding  chapters, 
because  in  either  case  it  is  a  function  of  one  variable 
alone. 

Case  I.  Suppose  that  y  has  the  constant  value  y0.  Then 
for  any  value  x0  of  x,  Az  =/(a?0  +  &x>  Vo)  —f(xo>  2A>)« 

Case  II.  Suppose  that  x  has  the  constant  value  x0.  Then 
for  any  value  y0  of  y,  Az  =f(x0,  y0  +  Ay)  -f(x0,  yQ). 

To  distinguish  between  the  Az's  in  the  above  cases,  we  shall 
use  the  notation  Axz  in  Case  I,  and  Ayz  in  Case  II. 
Prom  the  definition  of  a  derivative,  it  follows  that : 

—     is  the  derivative  of  z  with  respect  to  x  when 

Ax_ 

x  =  x0  and  y  is  constant ;  and 

is  the  derivative  of  z  with  respect  to  y  when 


limit 


limit 
Ay=0 


Ay 


y  =  y0  and  x  is  constant. 
126.    Definitions. 


limit 
A;r  =  0 


~AJ 

Ax 


is  called  the  partial  derivative 


of  z  with  respect  to  x  when  x  =  x0  and  y  is  constant.     It  is 
,  and  read,  "partial  —  d—x—  of —z  when  x=x0." 


written  — 
dx 


limit 
Ay=0 


Ay 


is  called  the  partial  derivative  of  z  with  respect 


•         •  Sz 

to  /  when  y  =  y0  and  x  is  constant.     It  is  written  — 

read,  "partial  —  d  —  y  —  of— z  when  y  =  y0." 


and 


y=y0 


In  general,  for  any  value  of  x,  ^inVt0 


value  of  y,  ^l 


A  z 

y 

Ay 


127.    To  interpret  Axz, 


_  dz 
By 

dx 


~4/ 

Ax 


dz 


Sz 
,  Ayz,  and  — 


=  — ,  and  for  any 
ox 


geometrically : 


Suppose  that  y  has  a  constant  value  yQ.     Let  A  be  a  point, 
distant  x0  from  the  2/z-plane,  on  the  curve  formed  by  the  inter- 


148       DIFFERENTIAL   AND  INTEGRAL   CALCULUS 


Y/ 


z 

A 

^\ 

F 

0 

/ 

X 

Fig.  45. 


section  of  the  plane  y  =  y0  with  the  surface  z=f(x,  y).     (See 

Fig.  45.)  From  A  draw  AB  per- 
pendicular to  the  ary-plane.  On 
the  line  of  intersection  of  the 
plane  y  =  y0  with  the  a^-plane,  lay- 
off BC  equal  to  Ax.  From  C  draw 
CD  perpendicular  to  the  xy--p\snae 
to  meet  the  curve  in  D.  From  A 
draw  AE  parallel  to  BC  to  meet 
CD  in  E. 

Then    Axz   is    geometrically    the 

distance   ED,   and    —  is    geo- 

dx  x-x 

metrically  the  slope  of  the  tangent  line  AE  to  the  curve  at 
the  point  A. 

Suppose  that  x  has  the  constant  value  x0.  Let  A  be  a  point 
on  the  plane  curve  formed  by  the  intersection  of  the  plane 
x=x0  with  the  surface  z=f(x,y). 
(See  Fig.  46.)  From  A  draw 
AB  perpendicular  to  the  xy- 
plane.  On  the  line  of  intersec- 
tion of  the  plane  x=x0  with  the 
#?/-plane,  lay  off  BG  equal  to 
Ay.  From  G  draw  GH  per- 
pendicular to  the  a;?/-plane  to 
meet  the  curve  in  H.  From  A 
draw  AK  parallel  to  B G  to  meet 
GH  in  K. 


dz 
Then  Ayz  is  geometrically  the  distance  KH,  and  — 


is 


geometrically  the  slope  of  the  tangent  line,  AS,  to  the  curve 
at  the  point  A.  - 


Since  —  and  —  are,  in  general,  func- 
dx  By 

tions  of  x  and  y,  we  may,  in  general,  repeat  the  operation  of 


128.   Let  2= /(a;,  y). 

f  x  and  y,  we  ] 
partial  differentiation  with  respect  to  either  x  or  y. 


TWO  INDEPENDENT   VARIABLES  149 

The  partial  derivative  of  —  with  respect  to  x  is  -r-f— \ 
2  dx  dx\dxj 


It  is  written  — -• 
oar 


dz       •  •       d   fdz\ 

The  partial  derivative  of  —  with  respect  to  y  is  —  [  —  ] 

d2Z  dv  dy  \dyJ 

It  is  written  — -  • 
dy2 

dz      .  •      d  f  dz 

The  partial  derivative  of  —  with  respect  to  x  is  —  ( — 

d2Z  dy  dx\ey. 

It  is  written 

dxdy 

Sz  d  /dz\ 

The  partial  derivative  of  —  with  respect  to  y  is  — (  —  J. 
Q2Z  ®x  dy  \dxj 

It  is  written 

dydx 

Similarly  for  higher  derivatives.     Thns  — •  ( — ( — ]  ]  is  the 

dx  \dy  \dyJJ 

partial  derivative  of  z,  first,  with  respect  to  y\  second,  with 

d3z 
respect  to  y ;  and  third,  with  respect  to  x.     It  is  written . 

dxdy 

129.  In  Chapter  XI,  in  the  equation  y  =  f(x),  where  f(x)  is 
a  function  of  one  variable  alone,  we  defined  dx  to  be  Aa?,  and 

dy  to  be  —  dx.     Then  in  z  =  f(x,  y),  in  Case  I,  dx  =  Ax  and 

dz  •  dz 

dz  =  —  dx,  and  in  Case  II,  dy  =  A?/,  and  dz  =  —  dy. 
dx  dy 

To  distinguish  between  the  dz's  in  the  above  cases,  we  shall 
use  the  notation  dxz  in  Case  I,  and  dyz  in  Case  II. 

130.  To  interpret  dxz  and  dyz  geometrically  : 

Produce  CD  to  meet  AF  in  F.  (See  Fig.  45.)  Then  dxz  is 
geometrically  the  distance  EF. 

Produce  GH  to  meet  ^4£  in  S.  (See  Pig.  46.)  Then  dyz  is 
geometrically  the  distance  KS. 

131.  So  far  we  have  considered  differentiation  only  in 
Cases  I  and  II.     It  remains  to  consider  Case  III. 

In  the  equation  z  =  f(x,  y)  where  x  and  y  both  vary  and 
are  independent  of  each  other,  let  x  have  the  value  x0  and 
y  the   value  y0.      Then  f(x0  +  Ax,   y0  -j-  Ay)  —  f(x0,  y0)  is  the 


150       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


increment  produced  in  f(x,  y)  by  giving  x  and  y  increments. 
Denote  it  by  Az. 

Then  Az  =  f(x0  +  Aa>,  y0  +  A2/)  -  /0»o,  2/o> 

132.  To  interpret  Az  geometrically  : 

Let  A,  with  coordinates  (x0,  y0,  z0),  be  a  point  on  the  surface 
z=f(x,  y).     Through  A,  pass  planes  parallel  to  the  xz  and  yz 

planes.  On  the  line  BC  of 
intersection  of  the  plane  ABC 
with  the  ory-plane,  lay  off 
BC=Ax.  On  the  line  .5(7 
of  intersection  of  the  plane 
ABG  with  the  #?/-plane,  lay 
off  BG  =  Ay.  Through  A 
pass  a  plane  parallel  to  the 
ar?/-plane.  Let  il^L,  the  line 
of  intersection  of  the  planes 
HGM  and  ECM,  meet  the 
plane  through  A  parallel  to 
the  ic?/-plane  in  N  and  the  surface  in  L.  Then  Az  is  geometri- 
cally the  distance  NL. 

133.  To  define  dz  when  x  and  y  vary  and  are  independent  of 
each  other : 

Let  A  with  coordinates 
(x0,  yQ,  z0)  be  a  point  on  the 
surface  z=f(x,  y).  Make 
the  same  construction  as  in 
Fig.  47.  (See  Fig.  48.)  Also, 
from  A  draw  a  tangent  plane 
to  the  surface  to  meet  the 
lines  drawn  perpendicular  to 
the  a?i/-plane  in  P,  F,  and  S. 

Since  Az  is  NL,  we  should 
wish  dz  to  be  NP.  Let  us 
define  dz  to  be  NP. 


TWO   INDEPENDENT   VARIABLES  151 

It  is  a  theorem  of  solid  geometry,  easy  of  proof,  that  if  a 
plane  cuts  a  rectangular  parallelopiped,  the  sum  of  two  oppo- 
site edges  cut  off  is  equal  to  the  sum  of  the  other  two  opposite 
edges.     Then  since  the  plane  AFPS  cuts  the  parallelopiped, 

JSrP  +  zevo  =  EF+KS. 

Now  JNP  =  dz,  EF=  dxz,  and  KS  =  dyz. 

.-.  dz  =  dxz  +  dyz. 

From  the  definitions  of  dxz  and  dyz,  it  follows  that 

dz            dz 
dz  =  —  dx-\ dy. 

dx  dy 

EXERCISES 

In  each  of  the  three  following  equations,  find  Axz,  Ayz,  _  an(^ 

dz       ,  ,  dx 

■ — ?  when  x  =  x0  and  y  =  y0: 

dy 

1 .  z  =  x2  —  3  xy  +  y2. 

Ans.   (2x0  —  3y0)Ax  +  Ax2;    (2  y0  —  3x0)Ay  +  Ay2; 

2x0-3y0;  2y0-3xQ. 

2.  z  =  3x2-2y2. 

Ans.  6  x0Ax  +  3  Ax  ;   4  y0Ay  —  2  Ay" ;  6x0;   —  4  y0. 

3.  z  =  xs  —  3  x2y  -f-  ys. 

Ans.  (3 x2  —  6 x0y0) Ax  +  (3  #0  +  3 ?/0) Ax  +  Ax'J ; 

(3y02-3x02)Ay  +  3y0Ay2  +  Ays-,  3x0  —  6x0y0;  3y02  —  3x02. 

In  each  of  the  four  following  equations,  find  dxz,  dyz,  and  dz : 

i  V  a        i  ydx       n  xdv 

4.  z  =  tan-1  —  Ans.  dxz  = f- — ,:  dvz  =  —0 — —n. 

x  x2  -f  y2 '     y       x2  -f-  y2 

,       ,       y  a        i  %ydx  2dx 

5.  z  =  log  tan  -•  ^4ns.  ax2  = ^y— ;  dz  = 


x  '     x  o   •    2y>     y  .    2v 

xr  sin  — -  x  sm  -^ 

^e-  x 

6.  2  =  e*y.  ^4ws.  dxz  —  y^dx ;  dy»  =  xexydy. 

7.  z  =  alog2/.       ^4?is.  dxz  =  x-1+losy\ogydx',  dyz  =  xlosylogx^- 

y 


152       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

In  each,  of  the  five  following  equations,  prove  that 

dh   =    d2z 
dy  dx      dx  dy 

8.  z  =  exy.  y 

11.    z  =  log  tan-- 

9.  z=(x2  +  y2y.  x 

.      ,H  N  12.    z='al0^. 

10.   z  =  x  log  (1  +  <ry). 

ox         •        dv 

13.  If  x=acos$,  y=a  smd,  prove  that  cos  8 \-  smO-Z  =  0. 

dO  dO 

14.  If  u  =  A  cos  (x  +  a£)  +  B  sin  (#  —  a£),  prove  that 

dhc  _    2  d2u  _  « 


15.  If  u  =  logV(#  —  a)2  +  (y  —  b)2,  where  x  —  a  and  ?/  —  b  are 

not  simultaneously  zero,  prove  that  — -  -\ =  0. 

dxr      dy 

16.  If  pv  =  a$,  where  a  is  a  constant,  find  dp  in  terms  of  dv 

and  ^  .         7  ^  7     ,  P  m 

An  s.  dp  = dv  +  sdv. 

1  v  0 

17.  If  u  =  log  (tan  x  +  tan  y),  prove  that 

sm  2  a h  sin  2y—  =  Z. 

dx  By 

18.  If  w  =  (5  £2  +  3  s2)*  prove  that  u5 -^j  +  «2^  +10  zt=Q. 

-ro  „    •       />  j_i    i.    9  5%  .      3?6  .  32?.t      A 

19.  If  u  =  rn  sm  n6,  prove  that  r  — -  +  r  —  +  — r  =  0. 

or         or      66- 


CHAPTER   XV 

INTEGKATION 

134.  In  Chapter  XI,  the  problems  discussed  were  of  this  sort : 
given  a  function  of  one  independent  variable,  to  find  its  differ- 
ential. We  shall  now  consider  the  converse  problem ;  namely, 
given  a  function  of  one  independent  variable  in  the  form  of  a 
differential  expression,  that  is,  consisting  of  a  function  multi- 
plied by  the  differential  of  the  variable,  to  find  the  function 
whose  differential  is  the  given  function. 

135.  It  is  evident  at  the  outset  that  if  we  can  find  one  func- 
tion whose  differential  is  the  given  function,  we  can  find  any 
number  of  such  functions,  because,  since  the  differential  of  a 
constant  is  zero,  functions  which  differ  only  in  their  constant 
terms  will  have  the  same  differential.  It  is  also  true  that 
functions  which  have  the  same  differential  can  differ  only  in 
their  constant  terms,  but  the  proof  of  this  theorem  would  be 
beyond  the  scope  of  this  book.  We  shall  assume  the  truth  of 
the  statement,  and  on  this  supposition  can  conclude  that  if  we 
can  find  a  function  whose  differential  is  the  given  function, 
this  function  plus  c,  where  c  is  an  arbitrary  constant,  will  con- 
tain all  functions  whose  differential  is  the  given  function. 

For  example,  2  x  dx  is  the  differential  of  x2  +  2,  x2  —  5,  or  in 
general,  x2  +  c,  where  c  is  any  constant,  and  x2  +  c  contains  all 
functions  whose  differential  is  2  a;  dx. 

136.  Definitions.  A  function  whose  differential  is  the  given 
function  is  called  an  indefinite  integral  of  the  given  function. 

Thus,  x2  +  2  is  an  indefinite  integral  of  2  x  dx. 

153 


154       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

The  process  of  rinding  an  indefinite  integral  of  a  function  is 
called  integration. 

The  constant  c  added  to  an  indefinite  integral  is  called  the 
arbitrary  constant  of  integration. 

Integration  is  denoted  by  the  symbol   j  . 

Thus,    I  2  x  dx  =  x2  +  c. 

The  symbol   j   is  read,  "  indefinite  integral  of." 

137.  Integration  is  the  inverse  of  differentiation.  Therefore 
to  integrate  a  function  we  must  reduce  it  to  a  form  where  it  is 
recognizable  as  the  differential  of  some  known  function.  The 
forms,  called  the  fundamental  forms  of  integration,  to  one  or 
other  of  which  we  shall  reduce  our  functions,  are  as  follows : 

/7,«+i 
un du  =  — he,  if  n  =£  —  1. 
11  +  1 

II.     I  —  =  log  u  +  c. 
J   u 

III.     Cau  du  =  -^L  +  c. 
J  log  a 

IV.  Ceu  du  =  eu  +  c. 

V.  (  cos  udu=  sin  u  +  c. 

VI.  I  sin  udu  =  —  cos  u  +  c. 

VII.  I  sec2  udu  =  tan  u  +  c. 

VIII.  I  cosec2  udu  =  —  cot  u  +  c. 

IX.  I  sec  u  tan  w  d^  =  sec  u  +  c. 

X.  I  cosec  u  cot  udu  =  —  cosec  it  +  c. 

XI.  I  tan  u  du  =  log  sec  it  +  c. 


INTEGRATION  155 

XII.     I  cot  udu  =  log  sin  u  +  c. 

XIII.     J  sec  m  du  =  log  (sec  %  +  tan  u)  -f-  c 
=  l°gtan(f  +  g+c. 

cosec  w  du  =  log  (cosec  % — cot  it)  +  c = log  tan  -  -f-  c. 


— =====  =  sin  J  -  +  c,  or  =  —  cos  *  -  +  c. 

-xAr2  —  ™.2  a  a 


XY 


XVI.     f-=^-=  =  -  tan-1  -  +  c,  or  =  -  -  cot"1  -  +  & 
*/  a24-ir     a  a  a  a 


Va2  —  n 

*    d"it 
a2  -\-u2     a 


XVII.     f — =  -  sec-1  -  +  c,  or  =  -  -  cosec"1  -  +  a 

J^W-a2     a  a  a  a 

f*    du           1    i       it —  a  .  1   i_  a —  u  .  n 

XVIII.        — -  =  — -log-— \-c,  or  =  — -log— — .+  & 

J  u2  —  a?     2  a        u  +  a  2  a        a  +  u 


XIX.     f      d^       =  log  0  +  VV  ±  a2)  +  c. 


V?t2  ±  a 

vv         C  dll  _ilt    , 

XX.     I  —  —vers     -  +  c. 

*^  a/2  ait  —  it2  a 

These  forms  are  not  fundamental  in  the  sense  that  no  one 
of  them  can  be  derived  from  one  or  more  of  the  others.  They 
contain  all  the  fundamental  forms  and  others  used  frequently 
in  practice. 

These  forms  can  be  readily  established  by  differentiation. 
For  example, 

</— 1—  un+l  +  c ) 

^  +  1 l  =  -^—(n  +  l)undu  =  undu. 

dx  n  + 1 

.-.    I  un  du  = +  c. 

J  n  + 1 


156       DIFFERENTIAL  AND  INTEGRAL    CALCULUS 

138.  The  following  theorems  are  of  constant  use  in  problems 
in  integration. 

Theorem  I.  A  constant  factor  may  be  written  either  before 
or  after  the  sign  of  integration. 

Theorem  II.  The  integral  of  an  algebraic  sum  of  two  func- 
tions in  the  differential  form  is  equal  to  the  sum  of  the  integrals 
of  these  functions. 

Proof  of  Theorem  I. 

Since  dcu  =  c  du,  where  c  is  a  constant,  theref ore,  from  the 

definition  of  an  integral,  „ 

cu=  I  c  du.  (1) 

Now   J  du  =  u,  by  applying  Form  I  where  n  =  0. 

.-.  c  I  du  =  cu.  (2) 

Equating  (1)  and  (2),  we  have 

J  c  du  =  c  I  du. 

Proof  of  Theorem  II. 

Since  d(u  +  v)  =  du  +  dv,  therefore,  from  the  definition  of  an 

integral,  „ 

I  (du  +  dv)  =  u  +  v.  (1) 

Now   J  du  =  u  and   I  dv=  v,  by  Form  I. 

Substitute  in  the  right-hand  member  of  (1). 

.*.    I  (du  +  dv)  =  I  du  +  I  dv. 

The  reasoning  in  Theorem  II  can  be  readily  extended  to  any 
number  of  functions.     It  can  therefore  be  shown  that 

J  (du+  dv-] h  div)=jdu+j  dv-\ hjdw. 


IN  TEGRA  TION  157 

139.    The  following  examples  will  illustrate  how  the  integral 
of  a  function  may  be  obtained  in  some  simple  cases. 

Example  1.   Find   C(xA  +  ar  —  3  x2  -f  2  a  —  fydx. 

C(tf _j_a;3_3a;2  +  2ic- 4)dx  =  Cx4dx  +  \x*dx  +  f  - 3 x2 dx 

+  I  2  x  dx  +  I  —&dx, 
by  Art.  138, 
=  I  x4  dx  +  I  x3  dx  —  3  I  x2  dx 

+  2  I  x  dx  —  4  I  dse, 
by  Theorem  I,  Art.  138, 

54  by  Form  I. 

Example  2.   Find  jT -1=  +  —  ]  d#. 


Va?      Vav 
rf-i=  +  -^=V^=  f^  +  f||,  by  Theorem  II,  Art.  138, 

=  I  x~*dx  +  I  x~y  dx 

=  2  a;  2  -|- 1  a;f  -f-  c,  by  Form  I, 

=  2-v^  +  f  ^  +  o. 

/x2—l 
dx. 
x 

(  — ^—  dx=  I  (  x )dx 

J         X  J   V  X 


=  J  x  dx  —  J  - 


a? 


=  |-  —  log  a;  +  c. 


158       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

EXERCISES 

Evaluate  the  following  integrals  and  in  each  case  show  by 
differentiation  that  your  result  is  correct. 

1 .  (  (xb  —  1)  dx.  13.  J  cos  (x  +  a)  dx. 

2.  C(x^-l)2dx.  14-  J  sin  (as -a)  cfo. 

3.  J  (Vx  +  Vx)2dx.  15,  J  x2  _  2 ' 

4.  f(Va?  +  3-v/^)cte.  16-  J^T^* 

J      x2  J  V 3  -  x 


3 

dx 


11.  J10z_2da;.  23.     jtan2icdic. 

12.  (V+1cto.  24.     fcot2a;cfo. 

The  arbitrary  constant  of  integration,  c,  can  always  be  deter- 
mined if  we  know  the  value  of  the  integral  for  some  particular 
value  of  the  variable ;  for,  since  a  constant  is  a  number  that 
has  the  same  value  for  all  values  of  the  variable,  it  is  known 
whenever  its  value  for  any  value  of  the  variable  is  known. 


CHAPTER   XVI 

PLANE  AKEAS  BY  INDEFINITE  INTEGKALS 
RECTANGULAR  COORDINATES 

140.  Let  y=f(x)  be  an  equation  in  which  f(x)  is  single 
valued  and  continuous  for  all  values  of  x  between  and  includ- 
ing two  values  a  and  b.  To  find  the  area  inclosed  by  the 
curve  y  =  f(x),  the  se-axis,  and  the  ordinates  corresponding  to 
the  abscissas  a  and  b  respectively. 

In  the  equation  y  =  f(x),  the  function  f(x)  may  be  positive 
or  negative  for  all  values  of  x  between  a  and  b,  or  it  may 
change  in  sign  any  number  of  times  as  x  increases  from  a  to  b. 

141.  At  first  suppose  that  f(x)  is  positive  for  all  values  of 
x  between  a  and  b. 

Under  this  supposition,  the  curve  y=f(x)  is  above  the  axis 


for  all  values  of  x  between  a  and  b.     Suppose  that  it  is  as  in 
Fig.  49. 

Let  OB  and  OC  represent  the  abscissas  a  and  b  respectively. 

159 


160       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


Let  OP  represent  an  abscissa  x,  a<x<b.  From  B,  P,  and  C, 
draw  ordinates  to  meet  the  curve  in  D,  S9  and  F  respectively. 

The  area  BDSP  is  obviously  a  function  of  x,  because  as  x 
assumes  different  values,  it  assumes  different  values.  Denote  it 
by  A.  Let  x  take  an  increment  Ax.  Then  A  takes  an  increment 
A  A.    In  the  figure,  PQ  represents  Ax,  and  PSRQ  represents  A  A. 

The  areas  of  the  rectangles  SQ  and  HP  are  fix)  Ax  and 

f(x  +  Ax)Ax  respectively.     From  the  figure  we  see  therefore 

that 

f(x)  Ax  <  A  A  <  f{x  +  Ax)  Ax, 

if  the  curve  is  rising  from  S  to  R,  or 

f{x  +  Ax)  Ax  <  A  J.  <  f(x)  Ax, 

if  the  curve  is  falling  from  S  to  R.     Divide  by  Ax. 

AA 
.\  f(x)  <  —  <f(%  +  Ax),  if  the  curve  is  rising  from  S  to  R, 

Hk,X 

AA 

or,  f(x  +  A#)  <  —  <  f(x),  if  the  curve  is  falling  from  S  to  R. 

As  Aa?  approaches  zero,  fix -\- Ax)  approaches  fix)  as  its 
limit.     Then,  whether  the  curve  is  rising  or  falling   from  S 

AA 
to  R,  has  a  value  between  fix)  and  an  expression  that 

Ax 
approaches  f(x)  as  its  limit  as  Ax  approaches  zero.     Therefore 

approaches  fix)  as  its  limit  as  Ax  approaches  zero.   That  is : 


^dA% 
dx 

.-.  —=f(x). 
dx     JK  } 

.\  dA=f(x)dx. 
.-.  A=  J  f(x)dx. 


But 


limit 
A:r  =  0 

LAx 

limit 

~AA~ 

Ax  =  0 

Ax 

PLANE  AREAS  BY  INDEFINITE  INTEGRALS      161 


Suppose  that  Cf(x)dx  is  <£(a)  +  c.  Then  the  area  inclosed 
by  the  given  curve,  the  sc-axis,  and  the  ordinates  correspond- 
ing to  the  abscissas  a  and  x  is  <£(a)  +  c,  where  c  is  a  constant 
at  present  undetermined.  To  determine  c:  Suppose  that  P 
moves  back  to  B.  Then  x  assumes  the  value  a.  Under  this 
supposition  the  required  area  is  zero,  and  <$>(x)  +  c  becomes 
<j>(a)  +  c.     .-.  c  =  —  <ft  (a). 

Therefore  the  area  inclosed  by  the  curve,  the  as-axis,  and 
the  ordinates  corresponding  to  the  abscissas  a  and  x,  is 
<f>(x)  —  <fi(a). 

Therefore  the  required  area  =  <£(&)  —  <£(a),  where 


i  f(x)  dx=<fi  (x)  -f  c. 


142.  Example.  Find  the  area  inclosed  by  the  curve  y  =  x2, 
the  ic-axis,  and  the  ordinates  corresponding  to  the  abscissas  1 
and  4  respectively. 

The  curve  is  as  in  Fig.  50. 

af  Ax  <  A^l  <  (x  4-  Ax)2  Ax. 
AA 

<{X-\-£±X)~. 

Y 


x2  < <(sc  +  Ax)2. 


Ax 

dec 

dJ.  =  x2dx. 


—  X2. 


^ 


=  I  x2dx  = 


af 


+  c 


Fig.  50. 


Let  P  move  back  to  B.     Then  x 
becomes  equal  to  1,    and  A  to  zero. 

xs  11 

.-.  A  —  —  -t-c  becomes  0=-  +  c.     .*.  c  =  —  -• 

Therefore  the  area  inclosed  by  the  given  curve,  the  sc-axis, 
and  the  ordinates  corresponding  to  the  abscissas  1  and  x  respec- 
tively is  xs     i 

3"~~3* 

Therefore  the  required  area  =  -^  —  i  =  21. 


M 


162       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


143.  Let  us  again  consider  the  problem  stated  in  Art.  140, 
and  suppose  now  that  f(x)  is  negative  for  all  values  of  x 
between  a  and  b. 

Under  this  supposition  the  curve  is  below  the  &-axis  for  all 
values  of  x  between  a  and  b.     Suppose  that  it  is  as  in  Fig.  51. 

Make  the  same  construction  as  in  Fig.  49,  Art.  141. 


Fig.  51. 


Since  f(x)  is  negative  for  all  values  of  x  between  a  and  5, 
the  areas  of  the  rectangles  UP  and  SQ  are  —  f(x-\-  Ax)  Ax 
and  —  f(x)Ax  respectively. 

Therefore      —  f(x  +  Ax)  Ax  <  A  A  <  —  f(x)  Ax 

if  the  curve  is  rising  from  S  to  B,  or 

— f(x)  Ax  <  A^4  <  —  f(x  +  Ax)  Ax 

if  the  curve  is  falling  from  S  to  R. 

Divide  by  Ax.     Then,  whether  the  curve  is  rising  or  falling, 

has  a  value  between  —f(x)  and  an  expression  which  ap- 

Ax 

proaches  —f(x)  as  its  limit  as  Ax  approaches  zero. .  Then,  as 
in  the  preceding  case, 

limit  TA^~|         /(  s 

.-.  A  —  —  \f(x)dx. 

Therefore    the     required    area  =  —  [<j>(b)  —  <£(«)],     where 
\f(x)dx=  cj>(x)  4-  c. 


PLANE  AREAS  BY  INDEFINITE  INTEGRALS      163 


144.  Example.  Find  the  area  inclosed  by  the  curve 
y  =  (x  —  1)  (x  —  2)  and  the  aj-axis. 

The  curve  is  as  in  Fig.  52. 

The  required  area  is  the  region  inclosed  by  the  part  of  the 
curve  between  x  =  1  and  x  =  2,  and  the  a>axis. 


...  A  =  -  C(x-l)(x-2) 
1  3        2 


cto 


+  c. 


When  x  =  1,  ^L  =  0,  and  therefore 
c=i-|+2=f 


Fig.  52. 


3    i    ^        -    i    6 

145.  Suppose  now  that  in  the  problem  stated  in  Art.  140, 
f(x)  changes  in  sign  as  x  increases  from  a  to  b. 

Under  this  supposition  the  curve  cuts  the  a>axis  for  some 
value  or  values  of  x  between  a  and  b. 

Since  there  are  different  expressions  for  area  according  as 
the  curve  is  above  or  below  the  a>axis,  it  follows  that  to  find 
the  area  in  this  case  we  must  find  the  area  separately  for  each 
of  the  regions  above  the  a?-axis  and  for  each  below  and  add  the 
results. 

146.  Example.  Find  the  area  inclosed  by  the  curve  y=sin  x, 
and  the  cc-axis  between  x  =  0  and  x  =  2  -k. 

The  curve  cuts  the  cc-axis  when  x  =  ir. 
For  the  region  above  the  co-axis, 

A=  I  sin xdx  =  —  cos x -f  c. 

Therefore  the  area  of  the  region  above  the  #-axis  =  2. 
For  the  region  below  the  cc-axis, 

A  =  —  f  sin  x  dx  =  cos  x  +  c. 

Therefore  the   area  of  the   region 
-j?  below  the  ;»-axis  =  2. 
Fig.  53.  Therefore  the  required  area  =  4. 


161       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

147.  From  Arts.  141  and  143,  we  see  that  when  f(x),  as  de- 
fined in  Art.  140,  does  not  change  in  sign  as  x  increases  from 

a  to  b,  <ji  (b)  —  <jy(a)}  where    I  f(x)  dx  =  <f>  (x)  +  c,  denotes  the 

area  or  the  negative  of  the  area  inclosed  by  the  curve  whose 
equation  is  y=f(x),  the  x-axis,  and  the  ordinates  correspond- 
ing to  the  abscissas  a  and  b  respectively.  This  raises  the 
question  as  to  what  it  denotes  when  f(x)  changes  in  sign  as 
x  increases  from  a  to  b. 

Suppose  that  f(x)  changes  in  sign  for  n  values  of  x  between 
a  and  b.  Let  the  values  be  c1}  c2,  c3,  •••,  cn,  taken  in  this  order 
as  x  increases  from  a  to  b. 

<j>(b)  -<j,(a)  =  ^(cj)  -  <f> (a)  +  cj> (c2)  -  <£(ci)  +  <Kcs)  -  <KC2> 

H \-<j>(b)-cj>  (cn),  identically, 

=  [<£  (c2)  -  <f>  (a)]  +  [*  (c2)  -  <t>  (cx)]  +  [<£  (c3)  -  <£  (c2)] 

+  -  +  l<f>(p)-<j>(cn)^ 

Now  each  expression  in  square  brackets  denotes  the  area  or 
the  negative  of  the  area  of  one  of  the  regions  inclosed  by  the 
curve  and  axis,  the  area  being  denoted  when  the  region  is 
above  the  sc-axis  and  the  negative  of  the  area  when  the  region 

is  below.      Therefore  <£  (6)  —  <£  (a),  where   J  f(x)  dx  =  $  (x)  -f  c, 

denotes  the  sum  of  the  areas  of  all  the  regions  inclosed  by  the 
curve  and  axis  above  the  a>axis,  minus  the  sum  of  the  areas  of 
all  the  regions  below. 

POLAR  COORDINATES 

148.  Let  r=/(0)  be  an  equation  in  which  f(6)  is  single 
valued  and  continuous  for  all  values  of  9  between  and  in- 
cluding two  values  a  and  /3.  To  find  the  area  inclosed  by 
the  curve  r  =  f(6),  and  the  radii  vectores  that  make  angles 
of  a  and  /?  respectively  with  the  initial  line. 


PLANE  ABEAS  BY  INDEFINITE  INTEGBALS      165 


149.   At  first  suppose  that  f($)  is  positive  for  all  values  of  0 
betweeu  a  arid  /?. 

Suppose  that  the  curve  r  =  f{6)  is  as  in  Fig.  54. 

Let  OB  and  OC  repre- 
sent the  radii  vectores 
that  make  angles  of  a 
and  /3  respectively  with 
the  initial  line.  Let  OP 
represent  a  radius  vec- 
tor that  makes  an  angle 
6  with  the  initial  line. 
Let  6  take  an  increment 
A0.  Let  OS  represent 
the  radius  vector  that 
makes  the  angle  0  +  A0 
with  the  initial  line. 

As  6  takes  an  increment  A0,  the  area  BOP,  or  A,  takes  an 
increment  AA.     Then  from  the  figure  we  see  that 

*{/(<?)  j2A6>  <  AA<  i\f(0  +  A0)j2A0, 

if  f(ff)  is  increasing  from  P  to  S,  or 

4{/(0  +  A0)J2A0  <  AA  <  i\f(0)}aM, 

if  f(6)  is  decreasing  from  P  to  S. 
Divide  by  A$. 

•••  i!/wi2<|f  <i\f(p+m: 

if  /(0)  is  increasing  from  P  to  S,  or 

i[/(«  +  A0)}'<||<*{/(tf)}«, 

if  /(0)  is  decreasing  from  P  to  S. 

As  A<9  approaches  zero,  \\f(6  +  AO)]2  approaches  \\f(6)\2 
as  its  limit.     Then,  whether  f(0)  is  increasing  or  decreasing 


166       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

A.A 

from  Pto  JS,  has  a  value  between  j\f(0)\2  and  an  expres- 

A0 
sion  that  approaches  \\f{&)\2  as  a  limit  as  A0  approaches  zero. 


That  is, 


limit 
A0  =  O 


~AA~ 

_A0_ 

dA 
dO 

.-.  A 


=umi2- 


=um\2. 

=  iflf(0)V 


dO 


where 


=  *  OS) -*(«), 
\$\f{0)YdO=4>{0)  +  c. 


150.  Next,  suppose  that  in  the  problem  stated  in  Art.  148 
f(6)  is  negative  for  all  values  of  0  between  a  and  f$,  or  changes 
in  sign  as  6  increases  from  a  to  (3. 

Since  f{6)  is  real,  ^\f(9)\2  is  positive  for  all  values  of  $ 
between  a  and  /?.     Then,  whether  f(0)  is  positive  or  negative, 

dA 

—  is  always  positive. 

dd 

09 


=  <t>(J3)-<t>(a), 
if  1/0)1***  =  <t>(9)+c. 


where 

151.   Example.      Find    the    area    inclosed    by    the    curve 

A  =  -  f  (1  -  cos  6)  dO 

+  c. 


=  a  VI  —  cos  0. 


a- 


—  sin 


A     When  0  =  0,      .4  =  0. 


c  =  0. 


a. 


When  0  =  2tt,  JL  =  ^- 


2tt-0 


=  7ra^. 


Fig.  55. 


Therefore  the  required  area  =  ttcl2. 


PLANE  AREAS  BY  INDEFINITE  INTEGRALS      167 

EXERCISES 

When  tables  are  necessary,  use  them  to  four  places. 

1.  Find  the  area  inclosed  by  the  parabola  y2  =  Sx  and  the 
double  ordinate  corresponding  to  the  abscissa  2.  Ans.  ^. 

x2  —  l 

2.  Find  the   area   inclosed   by  the   curve   y  = ,  the 

x 

a>axis,  and  the  ordinates  corresponding  to  the  abscissas  -J-  and 
1  respectively.  Ans.  0.318. 

3.  Find  the   area  inclosed   by  the   curve   y=  — — r,  the 

ce-axis,  and  the  ordinates  corresponding  to  the  abscissas  —  1 
and  +  1  respectively.  Ans.  0.549. 

4.  Show  that  the  area  cut  off  from  a  parabola  by  any  double 
ordinate  is  -|  the  area  of  the  circumscribing  rectangle. 

5 .  Find  the  area  inclosed  by  the  parabola  x1  +  y"1  =  a2  and 

the  coordinate  axes.  Ans.  — 

6 

6.  Find    the    area   inclosed    by   one    arch    of    the    curve 
y  =  cos  (  x  +  —  ),  and  the  a?-axis.  Ans.  2. 

7.  Find  the  area  inclosed  by  the  curve  y  =  —  ,  the 

V4  —  x2 
cc-axis,  and  the  ordinates  corresponding  to  the  abscissas  —  2 

and  +  2  respectively.  Ans 


IT. 


8.  Find  the   area   inclosed   by  the   semicubical   parabola 
ay2  =  xs,  the  y-axis,  and  the  line  y  =b.  Ans.   —  Vot62. 

In  each  of  the  seven  following  curves,  find  the  area  inclosed 
by  the  curve,  the  a?-axis,  and  the  ordinates  corresponding  to  the 
abscissas  set  opposite  the  equation. 

9.  y  =  x(x  —  1)  (x  —  2).     x  =  —  J,  x  =  S.  Ans.  3.141. 

10.  y=(x-1)(x-2).      x  =  ±x  =  l.  Ans.  0.261. 

x 

11.  xy  =  4:.  flj  =  l5a;  =  5.  Ans.  6.44. 

12.  y  =  sin  x.  x  =  1,  x  =  3.  Ans.  1,530. 


168       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

13.  y  =  cos  x.  x  =  2,x  =  l.  Ans.  3.566. 

14.  y  =  •  jc  =  1,  #  =  li  .4ns.  0.446. 

15.  ?/ =  10*.  a  =  ia  =  2.  4ns.  42.06. 

In  each  of  the  four  following  equations,  find  the  area  inclosed 
by  the  curve,  and  the  radii  vectores  corresponding  to  the  values 
of  0  set  opposite  the  equation. 

16.  r  =  tan  0.  0  =  0,0  =  --  Ans.  0.107. 

4 

17.  r  =  sini0  +  Cosi0.    0  =  0, 0  = j-  Ans.  0.539. 

18.  r  =  a0.  6  =  35°,  0  =  70°.  Ans.  0.266  a2. 

19.  r  =  e&.  0  =  -,0  =  --  Ans.  1.308. 

4  2 

Find  the  area  inclosed  by  each  of  the  following  curves : 

20.  r2  =  a2(l-cos0).  Ans.  (7r  +  2)a2. 

21.  r  =  a(l  —  cos  0).  Ans.  f  7ra2. 

C  6      \ 

Make  use  op   \  cos2  0  cW  =  -  +  -  sin  6  cos  6. 
J  2     2 

22.  r  =  a  sin  0.  Ans.  -a2. 

4 

23.  r  =  1  +  2  cos  0.     (Outer  curve.)  4ns.  2tt  +  |V3. 

24.  r  =  2  +  cos  0.  4ns.  f  tt. 

25.  ?-  =  a  +  &  cos  6,  \b\  >  |a|.     (Outer  curve.) 

4?2S. 


f  <>  i  &2\  &  i  3|a|    /To 5      i/i  _i      a        .   _t  v 62—  a2 

[a-  +  tt  J  0  H — It-1  V  6  —  a-,  where  6  =  cos  x  —  -  =  sm  x — ■ , 

\2J2  b  \b\ 

if  a  and  6  have  the  same  sign ;  (a2-\ — )  ( 7r  —  0  ]  H — ^  V62  —  a2, 

*     /y  ~\/ Jl     ft 

where  6  =  cos-1 =  sin-1— — ,  if  a  and  b  have  opposite 

b  \b\ 

signs.  '    i 

26.  r  =  a  +  6cos0,  |a|>|&|.  4ns.  (2  a2 +  &*)-. 

27.  r  =  a  +  6  cos  0,  \a\  =  |6|.     (See  21,  above.)        4ns  §  thx2. 


CHAPTER  XVII 

METHODS   OF  INTEGEATION 

In  Chapter  XV,  we  saw  how  to  obtain  the  indefinite  integral 
of  a  differential  expression  in  a  few  simple  cases.  In  this 
chapter,  we  shall  investigate  methods  whereby  the  integral  of 
more  complicated  expressions  can  be  fonnd. 

For  convenience  in  writing,  the  arbitrary  constant  will  be 
omitted  in  this  and  the  following  chapter. 

INTEGRATION  BY  SUBSTITUTION 

152.  In  many  problems  in  integration  the  integral  can  be 
evaluated  by  reducing  it  to  one  of  the  fundamental  forms  by 
means  of  the  substitution  of  a  new  variable.  The  following 
examples  will  illustrate  the  process. 

Example  1.     Find   f-^L. 
J  x  —  1 

The  differential  of  the  denominator  is  the  numerator.     The 

integral  will  therefore  reduce  to  Form  II  by  the  substitution  of 

u  for  x  —  1. 

Let         x-l  =  u.     .-.  dx  =  du,  and   f-^-=  f^L 

J  x  —  1     J    u 

Xow  C*Ol  -  log  u,  by  Form  II. 

J    u 


J_^_  =  iog  („_!). 


Example  2.     Find    f- 


-\Jl-\-x 

The  differential  of  the  expression  under  the  radical  in  the 
denominator  is  the  numerator.  The  integral  will  therefore 
reduce  to  Form  I  by  the  substitution  of  u  f or  1  +  x. 

169 


170       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

JjQtl  +  x  =  u.     .-.  dx=du,a,n&  |  —  =  1 — ^=  j  u~% 

J  Vl  -\-x     J  Vu     J 

Now  J  u~*  du  =  2  w*  by  Form  I. 

...    f^=  =  2(l  +  x)i. 
J  VI  -\-x 

Example  3.     Find   fsec2  (2  —  3  x)dx. 

This  integral  appears  as  if  it  may  reduce  to  Form  VII. 
Let  2  —  3  x  =  u. 

.-.  dx  =  — — ,  and    j  sec2 (2  —  3  x)dx  =  —  -J-  I  sec2  u  du. 

Now  —  i  |  sec2  u  du  =  —  \  tan  w,  by  Form  VII. 

.-.    fsec2  (2  -  3  x)dx  =  - 1  tan  (2  -  3  x). 

Example  4.     Find  j  ea+bxdx. 

This  integral  appears  as  if  it  may  reduce  to  Form  IV. 

Let      a  +  bx  =  it.     .-.  dx  =  — ,  and   j  ea+6a:c&£  =  -  (  eu  du. 

b'         J  bJ 

i  r         l 

Now  -  I  eudu  =  -eu,  by  Form  IV. 

bJ  b        J 

"J  6 


Example  5.     Find      j  - 


ax2  +  5a?  +  c 

The  denominator  is  rational  and  of  the  second  degree.  The 
integral  should  therefore  reduce  to  Form  XVI  or  XVIII. 
Divide  both  numerator  and  denominator  by  a. 

dx 


/dx  _1  C 

ax2  +  bx  4-  c     a  J   ^ 


,   b      .  c 
v?  +  -x-\-  — 
a        a 


METHODS   OF  INTEGRATION  171 


Now  a*  +  Zx  +  ±  =  a?  +  »x  +   £-    +--JT T2 


a        a  a        \2aJ      a     4:0? 

&  V  ,  4  ac  -  62 

^  +  777.    + 


2aJ  4,0?    ' 

by     b2  _  4  ac 
or  =  [x  +  —  ] J--J — 


2  a  la? 


/dx  _1  /» dx 

ax2  +  bx  +  c~aj  /         6  V  |  4ac-?>2? 
\       2  ay  4  a2 

1  /»  dx 


2  a  4:0? 


Let                       a?  H =  w.     .*.  dx  =  duj 

2a 

■j                 r         dx          _  1  /»  a^ 

J  ax2  +  &x  +  c  ~  a  J     n  ,  4  ac—b2' 


or 


a  J 


4  a2 
&2  —  4  ac 


4  a2 
Suppose  that  62  —  4  ac  is  negative. 


Then  4  ac  —  62  is  positive,  V4  ac  —  62  is  real,  and 

f ^ ,  =        2a         tan-     t    U         ,  by  Form  XVI. 

J     ,     4  ac  -  52      V4ac-62  V4ac-fr2 

4a2  2  a 

xH 

/»        dx         _1         2a  j  2a 


ax2  +  &x  +  c     a  ^4  ac  —  b2  V4  ac  —  52 

2a 

2  .     _j    2ax  +  &_ 


tan 


V4  ac  —  62  V4  ac  —  b2 

if  &2  —  4  ac  is  negative. 


172       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Suppose  that  b2  —  4  ac  is  positive. 

Then  V&2  —  4  ac  is  real,  and  r— — 

V  5-  —  4  ac 
u 

f         du =  1         2a        log ,  2a       ,  by  Form 

J  u2  |  &2-4ac      a2Vb2-±ac        u  |  Vfr2-4ac     XV1II. 
4  a2  2  a 

6  "\      V&2  —  4  ac 


dec  2  a  V        2  a)  2  a 


/dx         __         la         1 
ax2  +  bx  +  c~  % ^/W^tac  °g 


b  \  .  V&2  —  4  ac 

X+7TZ  1  + 


2a  2a 


1      2  a%  +  5  —  V&2  —  4  ac 


V&2  —  4  ac        2  a#  +  b  +  V&2  —  4  ac 

if  &2  —  4  ac  is  positive. 

Suppose  that  62  —  4  ac  =  0. 

1  T  aw  1  C  du  , 

Then  either  -  I  -A r2  or  -  I  r — j—  becomes 

a  J     o  ,  4  ac  —  er        a  J     9      cr  —  4  ac 


tr  H — —  w 

4  a2  4  a' 


±  C^H  which,  by  Form  I,  reduces  to 

aJ  u2  a  u 

dx  1       4 


'  J  ax2-{-bx-\-c         a  x  ■     b 

2a 


2ax-\-b 


EXERCISES 


if  b2  -  4  ac  =  0. 


Show  by  integration  that : 

dx  1  1      4  x  —  3 


1 


r   dx    =j_1   : 

Jl6x2-9      24     &4z  +  3 

>      d*       =i,tan-ig^. 
J9a2  +  16      12  4 

.     f dx =  1  log  (4  x  +  VI  +  16  ft2)- 


VTTl6a;2     4 


METHODS   OF  INTEGRATION  173 


4.  I —  =  -  sin1 4  x. 
J  Vl-16a2     4 

5.  I =  vers    -• 

J  VSx-x2  4 

6.  C(2  +  3  ®)2cfa  =  1  (2  +  3  x)\ 

9.   JVa  +  6o5 cfcc  =  —  V(a  -j-  6x)3. 

io.    CxV^Tbx-dx  =  -2(2a-Sbx)^(^±W. 

J  15  b2 

11  C  C*X  —  ~^3  —  4  37 

J  V3  —  4  a;  2 

12.  f(ex  +  e-B)sdo;  =  -  (e2x  +  4  a,  -  e~2x). 

1 3 .  f(eax  +  e~ax)  dx  =  -  (eax  —  e~ax) . 
•/  a 

14-  /S=leto+le&+l^+e*+log(e"-1>- 

15.     |  (sin  3 g  +  cos 3  x)  dx  =  |  (—  cos  3  a;  +  sin  3 a). 
16'  f^E?*°  =  ]og(!"  +  *>*<»-*)*■ 

J  V2a2+3o;-3    V2       V  2V27 

19.     f         ■.*»  =-Lm^i£ng. 

J  V3  +  3  a;  -  2  a2      V2  V33 


174       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


20.   / 


dx 


3_iog4^3-V41 


2^  +  3^-4      V41       4a  +  3+V41 


21      f  ^  =  _1 

22.      f  <*" 


tan 


_!  3  a?  —  1 


-  =  - — =  tan  *  — — — 
F  +  2x  +  3      V2  V2 


23 


24 


da? 


V2iB2+3a?+l      V2 


—  log  (4  x+3+2V2V2a;2+3aj+l). 


•/ 


dx 


s/ax2-{-bx-\-c      V 


log  (2ax+b-\-2  -Va^ax2  -\-bx-\-c), 

if  a  >  0. 


V—  a 


•   _i  —  2  ax  —  b     • «        .  n 
sin    —  ■  — ,    it  a  <  0. 


V  &2  —  4  ac 


25 


/ 


icdaj 


Suggestion. 


Va#2  -f-  bx  +  c 


=  -  Va£2  +  te  +  c  —  t^—  I  - 
a  £  U-J 


dx 


2  ax-\-  b 


^fax2-\-bx-\-c 
b  1 


Vax2  +  bx  +  c     %  a  Vax2  +  bx  +  c     2  a  vW2  +  &x  +  c 
identically. 

26-     f-|^=J-^  =  log[(^  +  3)2(^-2)]. 
J  x-  +  x  —  b 


+ 

f  gjk    ==_la,_ A  l0g(3-4aj). 
J  3-±x         4        16  ■  y 

28.     f  ^^    =  - x2  -  -  x  +  —  log  (3  x  +  2). 
J  3a; +  2      6         Q        97    5  V 


29 


•/ 


+  2  9        27 


V2^=-iC2V2-7^tV(2_7*)3]- 


•a/  litt/ 


80./- 

31.      f 

J   Vl  +  a;      5 

J  a  +  bx      b 


.  V(3  +  2oj)3  -  6  V(3  +  2  jb) 


V3  +  2  x     4 

^^  =  |  V(T+^  - 1  V(TT^3  +  2  vr  + 
VI  +  a;      5  6 


x. 


33 


34 


35 


METHODS   OF  INTEGRATION 
x  dx       1 1 cc2  —  1 


175 


J  x'-l     4     5^  +  l 

r      dx  l 

^  a,V2-5a?      V2~    JV2^5^-  +  V2 


1    ,      V2-5x-V2 
-log 


/ 


i  _  _  /  Va  +  &ic  —  Va 


a?  Va  +  6a?     Va        \Va  +  &»  +  Va 

2 


,  if  a>0. 


tan" 


37 


•/ 


V—  a 
da?  1 


»Ja±to   ,f  a  <  a 
*    —  a 


x^x2-3x+7         V7 
1 


,Va2-3a+7  +  V7       3   \ 

— m- 


X 


Suggestion.     Let  x  =- 
u 


38 


39 


■■/- 


dx 


.   _l{3x-2\ 

—  sin    a  l  !• 


sm" 


aV4a2  +  3a;-l 


5a; 


1  ,      /Vaa?2+&a?-hc-f  Vc  .      & 
= -log 


X 


if  c>0. 


40 


41 


42 


ajVaa^+to+c         Vc 

1        -  -\f    bx  +  2  c    \    -o  M   .  n 

=  — — —  sm  *   — — ==  ),  it  c  <  0 

V—  c  \a?  V&2  —  4  ac/ 

da?  2  Va  +  6a? 


2Vc/ 


J  ax2-\-bx-\-c 

/dx      r 
m~2   /y»¥ 


Va  +  bx  v 

=  ilog(a^+te+c)-Ar 
2  a  2aJ 


dx 


ax2+bx-\-c 


x2  —  x3 


f  +  f  +  **  +  log(**-l)J. 


Suggestion.    Let  cce  =  ?^. 
_   i 


43 


/ 


dx 


5  3 

a^  +  a?* 


=  8 


—  a?8  +  log  (a?8  +  1) 


176       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

X2        X*    .    X        X*    .    X2  l    .    ,         .    l    ,         ' 

J— g-  +  S-3-  +  ¥-^+l0g(^  +  l) 


44.     C    xc^x    —  1 


45. 


/ 


3. 

x2dx 


46 


■/i 


2a  — a; 
da? 


-2 


r-      1 

x2 


— [-2  a^2  +  V 2  a2  log 

3  a*  +  V2  a.. 


# 


2  +  5  sin  a?      V21 

C        dx 
Suggestion.      \ 


2tan-  +  5-V21 

1     !  2 

log 


x 


2tan^  + 5+V21 

Li 


2  +  5  sin  x     J  2  / 


(7x 


2  (  sin2  -  +  cos2  -^  +  10  sin  -  cos  - 
V       2  2J  2 


sec2  -  dx 


47 


48 


49 


■/ 


dx 


2  tan2- +  10  tan- +  2 

2  2 


V7tan'T-V3 

1     !  2 

log- 


Let  tan  -  =  u. 
2 


2  -5  cos  x      V21        V7tan?  +  V3 

Ll 


X 


dx 


■Sf= 


5  +  2  sin  a;     -y^l 
dx  2 


5  tan- +  2 

2     ,     _!  2 

tan  * 


V21 


2  cos  a;     V21 


tan' 


-'(SIN) 


RATIONAL  FRACTIONS 

153.  A  rational  fraction  in  £c  multiplied  by  dx  can  be  inte- 
grated, whenever  the  denominator  can  be  factored  into  linear 
or  quadratic  factors,  by  breaking  the  fraction  up  into  partial 
fractions. 

dx 


Example.    Find 


Let 


/ 


(x-iy(x-2) 

A 


+ 


B 


C 


(x-iy(x-2)~  (a? -I)3      x-1      x-2 


METHODS   OF  INTEGRATION  177 

Clear  of  fractions. 

.-.  l=A(x-2)+  B(x  -  l)(x  -  2)  4-  C(x  -  l)2. 

Equate  coefficients. 

.-.  B+C=0. 

A-3B-2C  =  0. 

-2A  +  2B+C=1. 

By  solving  the  equations,  we  get  A  =  —  1,B  =  —  1,  (7  =  1. 

1  _JL 1_      _1_ 

"  (x-l)2(x-2)~       (x-1)2      x-1      x-2* 

r        dx         _     r    dx        r  dx       r  dx 

"  J  (x-lWx-2)         J  (x-1)2     J  x-1     J  a 


(x-l)2(x-2)         J  (x-1)2     J  x-1     J  x-2 

1       .  ,      x-2 
-4-  log- 


ic— 1  x  —  1 

EXERCISES 

Show  by  integration  that : 
-       C      2x~1         7        i      (x-2)3 

lB  j  (x-l)(x-2)dx=l^-x~^l' 
J  (x  4-  1  Vx  4-  3Vx  4-  o)      8     8 


(x4-l)(a  +  3)(x4-5)      8     °  (a;  +  5)5(x  4-  1) 

_      f      2x3  +  l        .         .<.-',!      (x4-2)15 
3.     I- — -J j-  cfcc  .=  x2  —  6  a?  +  log  ^— - — ^-« 

J  (x  +  l)(x  +  2)  5    x  +  1 

4  1  'IX  -i  X 

J  x(x2  + 1)  Vx24-1 

J  (a?  +  Z)(ar  +  l)  5  5 

f       ft24-x  2__      11        (x-1)2      2  * 

6*   J  (x-l)2(x24-4)  6(aJ-l)  +  601Og"^4T  +  26tan    2" 

„      C xdx _1|      a;2  4-2x4-3       1         _iX4-l 

'   J(x4-l)(*24-2x4-3)~4     g      (x4-l)2     +V2  "vT 

_       r   dx        1,        (x4-l)2     .     1    ,      _j2x-l 

8.     I  -TTT  =  ^lQg    2  /.,+  — tan  * — - ■ 

J  x' 4- 1      6       r-«  +  l     -x/q  -x/o 


178       DIFFERENTIAL  AND  INTEGRAL    CALCULUS 

INTEGRATION   BY  PARTS 
154.   In  Art.  99  we  saw  that 

d(uv)  =  udv  +  v  du. 
Transpose.  .-.  udv  =  d(uv)  —  v  du. 

.-.    I  udv=  i  d(uv)  —  I  vdu. 
Now  J  d(uv)  =  uv. 

.:    j  u  dv  =  uv  —  I  v  du.  (1) 

The  use  of  (1)  is  called  integration  by  parts.      By  its  aid, 
the  desired  integral  can  often  be  found. 

Example  1.    Find   j  sm2xdx. 

Let  u  =  sin  x,  and  dv  =  sin  x  dx. 

.\  du  =  cos  x  dx,  and  v  =  —  cos  x. 

Substitute  in  (1). 

.•.    j  sin2  x  dx  =  —  sin  x  cos  x  -f  j  cos2  x  dx.  (2) 

Now  cos2  x  =  1  —  sin2  x. 

.'.    I  cos2xdx=  I  dx—  I  sin2£ccfcc 

=  a;  —  I  sin2  x  dx. 
Substitute  in  (2). 

.-.  sin2  xdx  =  —  sin  x  cos  x  +  x  —  I  sin2  #  cfcc. 
Transpose   j  sin2  aj  dx,  and  add. 

.'.  2  I  sin2  x  dx  =  —  sin  #  cos  #  +  x. 

.'.    I  sin2#c?#  =  —  -since cos #  +  -. 
%)  2  2 


METHODS   OF  INTEGRATION 


179 


Example  2.     Find   j  eax  sin  nx  dx 
Let 


u  =  eax,  and  dv  =  sin  nx  dx. 


.-.  die  =  aeax  dx,  and  -u  = cos  nx. 

Substitute  in  (1). 

/eax  sin  nx  dx  = eax  cos  nx  -f-  -  I  eai  cos  nx  dx.       (2) 
?i  n  J 

In    I  eazcos  nx  dx,  let  w  =  eax,  and  c?v  =cos  no?  dx. 


\  du  =  aeax,  and  v  =  -  sin  nx. 

n 


Substitute  in  (1). 


■■/ 


eax  cos  nx  dx  =  -  eax  sin  nx 
n 


--f 

nJ 


eax  sm  no?  dec. 


(3) 


Substitute  (3)  in  (2). 

C      .         ,  1  a 

I  ea35sinwa5c?aj=  —  ea*cosn#+- 

4/  ft  71 


ft 


e"sinwa! 


aeax  sin  w#  —  neax  cos  no? 


»r 


nJ 
a2  C 


eax sin  nxdx 


j 


eax  sin  nx  dx. 


a2  C 

Transpose I  eax  sin  nx  dx,  and  add 

nl  J 

.'.  (1+—  j  f  eax  sin.  nx  dx  = 


aeax  sin  nx  —  neax  cos  nx 


-!■ 


eax  sin  nx  dx 


w 


aeax  sm  nx  —  neax  cos  nx 


a2  -\-w 


EXERCISES 

Show  by  integration  that : 


x2 


1.     j  x\ogxdx  =  — 

■s 

af  log  #  cZa?  = - 

m  +  1 


logo;  — 


2.     j  x2  logx dx  =  — 


2 


logo; 

8        3 


logo; — 

m  +  1 


180       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

4 .  Cx  sin  a?  dx  =  sin  x  —  x  cos  x. 

5 .  Cx2  smxdx  =  2x  sin  x  —  (x-  -  2)  cos  x. 

6.  Cx2  cos  acta  =  2  a;  cos  a;  +  (^2  -  2)  sin  a> 

a;eaida;  =  — (aa;  —  1). 
or 


~rf_2x     _2" 
a       a2      a3 


8.  lx2eaxdx  =  el 

9.  J  sin-1  a?  da?  =  a;  sin-1  a;  +  Vl  —  ^ 

10.  |  cos_1a;<:7a;  =  a;cos_1a;  — Vl— ^2- 

11.  ftan^a;  da;  =  x  tan^a;  -  -  log  (1  +  ^2). 

12.  I  log  x dx  ==  xlog x  —  x. 

r  1   •  1 

13.  I  cos2a;c7a;  =  -sma;cosa;  +  -a;. 

14.  (sin  x  cos  x  dx  =  -  sin2  a\ 

15.  j  sinn  a;  cos  a;  da;  = 

16.  I  sin  a;  cos"  a;  da;  =  — 

17.  J  sin  nx  cos  ma;  da;  = 


sinn+1a; 
n-f  1 

cosw+1a; 


n  +  1 

cos  (n  —  m)a;      cos  (n  +  m^ 
2(rc  — m)  2(%  +  m) 


sin  (n  —  m)x      sin  (n  +  ™)x 
ismnxsmmxdx=    2(n_m)'  2^Tmj~ 


r  -,        sin  (n  —  m)x      sin  (n  +  m)as 

19.   J  cos  na  cos  ma  rite  =  _  m)    +    2(n  +  m) 


■J' 


20.     i  eax  cos  wa;  da;  = 


qeaa:cos  ft  a;  +  we"*  sin  ?ia; 
a2  +  ?i2 


CHAPTER   XVIII 

SEDUCTION   POEMULAS 

ALGEBRAIC   REDUCTION   FORMULAS 
155.    These  are  formulas  whereby  the  integral 

Cxm(a  +  bxn)pdx 

may  be  made  to  depend  upon  a  similar  integral  with  a  more 
convenient  value  for  m  or  p,  or,  in  some  cases,  may  be  com- 
pletely evaluated.     They  are : 

J  afn(a-{-bxnydx 

=  *"-+>  + toy* _  (m-n  +  l)«   r  _^       hxyd      (A) 
(np  +  m  +  l)b         (np  +  m  +  T)bJ  '  v    J 

Cxm(a-\-bxn)pdx 

=  ^K«  +  tory  +      anp       r         bxny-ld  (B) 

np  +  m-\-l        np  +  m  +  lJ 

Cxm(a+bxn)pdx 

(m  +  l)a  (m  +  l)a        J  V  ;       ,KJ 

Cxm(ci  +  bxnydx 

xm+1(a  +  bxn)p+1     np  +  m  +  n  +  1  r  m,     .  ,  _x _ , ,  7      , ^N 
n(_p  +  l)a  (>  +  l)a    J      v  y  '  v    J 

Formula  (A)  is  used  to  reduce  the  exponent  m,  (B)  to  reduce 
the  exponent  p,  (C)  to  increase  the  exponent  m,  and  (D)  to 
increase  the  exponent  p.  Formula  (O)  or  (Z>)  is  used  when  m 
or  jp  is  negative. 

181 


182       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

156.    Derivation  of  Formulas. 
To  derive  (A) : 

Integrate  by  parts  where  xm~n+1  =  u, 

u  =  xm~n+1,  and  dv  —  (a  +  bxn)pxn-1dx. 

.-.  du  =  (m  —  n  + 1)  xm~n,  and  v  =  ^—^ l-—- 

J  nb(p  +  l) 

.-.   J  xm(a  +  bxn)pdx 

xm-n+Xa  +  6a.r  1  _  m_n+1     /•  Baf^cfa.        (1) 

Now 
|  ajm_w(a  +  bxnY+1dx  =  Cxm~n(a  +  &jcn)  (a  +  te^cfa 

=  a  C xm-n(a  +  6ajny  da;  +  6  J  a;m(a  +  bxn)pdx. 

Substitute  in  (1),  transpose,  add,  and  divide  by  W-f. — ±_. 
w  w(p  +  1) 

.-.   Cxm(a  +  bxn)pdx 

=  x™-^\a  +  bxy^ _  (m-n  +  l)a    jV-(a  +  hxnydXy 
(np  +  m  + 1)  b         {np  +  m  -f- 1)  bJ 

which  is  formula  (^4). 

To  derive  (C)  : 

( Tfb n  ~\~  1 1  a 

Transpose  the  integrals  in  (A)  and  divide  by  -) ±—  • 

ox/  (wp  +  m+l)& 

.-.   r af*-"(a  +  bxn)pdx 

=  x™-^(a  +  bx«)^ _  (np  +  m  +  l)6  T  m(a  +  &af ^ ^ 

(m  —  n  +  1)  a  (m  -?i  +  l)aJ 

;Let  m  _  n  =  m',  or  m  =  w+m',  substitute  and  drop  the  accent. 

.-.   C xm(a  +  bxnydx 

=  ar+^a  +  ^r1  _  (np  +  m  +  ^  +  l)&  (V+»(a  +  bxnydXf 
(m  + 1)  a  (m  +  1)  a         J 

which  is  formula  (O). 


REDUCTION  FORMULAS  183 

To  derive  (B)  : 

Integrate  by  parts  where  (a  +  bxn)p  =  u. 

u  =  (a  +  bxn)p,  and  dv  =  xmdx. 


:  du  =  bnpxn~\a  +  bx'y^dx,  and  v  = 


x 


m+l 


m  +  1 


.-.   C xm(a  +  bxn)pdx 

=  *+\a  +  b*y _ Jmp_  rm+n(a  +  bxny-hlx.         (1) 
m  +  l  m  + U 

Now 

pf  (a  +  bxn)pdx  =  f#m(a  +  tow)  (a  +  bxn)p~ldx 

=  a  J  a;m(a  +  boffy^dx  +  6  pcm+n(a  +  &scn)p_1da;. 
.-.   Ja;",+"(a+6aJn)p-1cfoj=l  pz;w(a  +  &a;n)*da5— -  fxm(a+bxny-1dx. 

Substitute  in  (1),  transpose,  add,  and  divide  by    *     — - — . 

.-.   \xm(a  +  bxn)pdx 

np  -\-m  +  1        np-{-m-\-U 
which  is  formula  (JB). 

To  derive  (Z?) : 

Transpose  the  integrals  in  (B)  and  divide  by — 

np  -f-  m  + 1 

.-.   p5m(a  +  &a?l)p-1cfa; 

=  _  «f*'(«  +  6*y  +  np  +  m  +  1  r ,(fl  +  ^^ 
anp  anp       J 

Let  _p  —  l=p'  or  p  =  1  +!>'>  substitute  and  drop  the  accent. 

.-.   Cxm(a  +  bxn)pdx 

xm+\a  +  &ajnV+1  .  np  4-  m  +  n  + 1  /•  m/     ,   ,   _._. ,  7 

= ) —    —  y h  -^b ^r— J —  I  xm(a  +  6a3ny+1cZa;, 

n(p  +  l)a  n(p  +  l)a    J      v  y  ' 

which  is  formula  (D). 

These  formulas  fail  when  a  denominator  becomes  zero. 


181       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

157.    As  an  illustration  of  the  method  of  application  of  these 
formulas,  consider  the  following  example. 

x2dx 


Example.     Find 


/: 


VV- 


,2        ™2 


By  an  application  of  formula  (A)  this  integral  may  be  re- 
duced to  an  expression  containing  |  —  ,  and  can  therefore 
be  integrated.                                         Va2— x2 

f    x\dx     =  Cx\a2 -  x2)-*  dx. 
J  Va2  —  x2     J 

Apply  formula  (A),     m  =  2,  n  =  2,  p  =  —  A,  a  =  a2,  b  =  —  1. 

/'  9/  o       ON-l,  x(a2  —  x2)*  ,  a2  /*      da; 


2^  Va2  —  x2 


x-\/a2  —  x2  ,  a2   ■     ,  # 
H —  sin  1 


2  2  a 

2  cZ#              a?  Va2  —  x2      a2   •   _t  cc 
= h  -p:  sin  x  -  ■ 

Va2  —  x2  2  2  a 


/x2cZa; 


EXERCISES 

Show  by  integration  that : 

V«2  —  x2dx  =  ~  Va2  —  x2  H —  sin-1  -  • 
2  2  -a 

2  

Va2  +  x2  dx  =  -  Va2  -f-  cc2  +  —  log  (x  -f  Va2  +  x2). 
dx  x  3  a2  —  2  x2 


(a2-^)!     3a2(a2-^)t  a* 


4.  f aVa2  -  x2-cto  =  -  1  V(a2  -  x2)3. 

5.  CxW^^dx=-(a2  +  2x2)V'^^--log(x+V¥^2). 
«y  8  8 

6.  Cx2V^F^xT2dx=  -^VCa^^+^Va^^+^sin-1-). 


REDUCTION  FORMULAS  185 

/x2  dx         x    .— a2  ,— rx 

,  a       2  =  o  V x2  ±  a2  T  2"  loS  0&  +  Var  ±  av- 


r&         _       Va2  —  aj2 
ar9Va2-ar9~  tfx 


9 .    f  (a2  _  ^2)f  ^  =  I  (5  a2  _  2  ^)  Va2  -  x2  +  ^  sin-1 

.0.    I  —        ■        =  —  V2  aa?  —  x2  +  a  vers-1  -  • 
J  V2ax-  x2  « 


dx  V2  ax  —  x2 


%J     nt- 


x-y/2  ax  —  x2  ax 

0      C        x2  dx  _  f  x       3  b  \     

LZ'    I  —    o ~a — 9  V aa;2  +  bx  +  c 

J  ^ax2  +  bx  +  c      V2a      4a7 

,  3  62  —  4  ac  /~  cfa 


If: 


8  a-    'J  Vax2  +■  ftsc  +  c 

Suggestion.    ax2+&£+c=a[(  x  +  —  )  —     ~4qc1.     Let  w  =  z+ 

LV       2a/  4a3     J 


TRIGONOMETRIC   REDUCTION   FORMULAS 

158.    The  following  are  reduction  formulas  for  trigonometric 
functions : 

j  tann  x  dx  = j  tanw  2  x  dx.  (1) 

f cot"  x  dx  =  -  cotW"1  -  -  foot'1"2  x  dx.  (2) 

/.   _          _      7           sinm_1  a;  cosn+1  a;  .  m  — 1  /•  .   m_0  _      7 

sm™  oj  cosw  x  ax= 1 I  smM  2  x  cosn  a?  $x. 
m-\-n              m-j-nj  /QN 

j  sinmajcosna;<7a;= — 1 I  smmxcosn~2xdx  (4) 

J  m-\-n  m+nJ 

/cosw  x  dx  _   _        cosw+1  x  yi  —  m  +  2  /*cosn  a?  c7a?  ,~ 

sinm  x  (in  —  1)  sin™-1  x         m  —  1    J  sin  m~2  a? 

/cos"  x  dx  _         cosn_1  a?  n  —  1   /*cosw~2  a;  (7a;  ,«. 

sinm  a;        (??,  —  m)  sinw_1  a?      n  —  ww      sinm  x 


186       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

/sin"1-1  x  cos  x  ,  m  —  ir.  m_<>     ,  /r7N 

smm  x  dx  =  —  - 1 I  smw  2  x  dx.  (7) 

m  m    J 

/cosw_1  x  sin  x  ,  n  —  1  (*      n_2      -j  /QX 

cos"  x  dx=  — I |  cos"  2  x  dx.  (8) 

n  n    J 

/dx cos  a;  m  —  2  r     dx     _  ,<^ 

sinw  a;  ~       (m  —  1)  sin™-1  a;     m  —  1 J  sin"1-2  a; 

/cZa;    _  sin  a;  n  —  2Tdx  qqn 

cos"  a;  ~~  (w  —  1)  cosn_1  x     n  —  lJ  cosn_2  x 

159.    Derivation  of  some  of  formulas. 
To  derive  (i)  : 

J  tann x dx  =  j  tann~2 a? tan2a? dx 

=  J  tann-2  x  (sec2  a;  —  1)  dx 

=  |  tann_2 a;  sec2 x  dx—  I  tan"-2 a; efte 

tan"-1  a;  _  Aann-2  ^  ^ 
n-1       J 

To  derive  (3) : 

Integrate  by  parts  where  sinTO  xx  —  u. 

u  =  sin"1-1  x,  and  dv  =  sin  x  cosn  x  dx. 


.-.  du  =  (m  —  1)  cosm_2  x,  and  v  =  — 


cosn+1  X 


w  +  1 

.*.  j  sin"1  a?  cos71  a;  da; 

=  _  sin-1  a;  cos"*1  a;      m-1  fV^.,  ^  cogn-2  ^  ^ 
w  +  1  n  +  U 

Now 

Am™-2  x  cosn+2^c  dx  =  J  sinm"2  a;  cosn  a:  cos2  x  dx 

=  (  sin"1-2  x  cosw  a;  (1  —  sin2  a;)  dx 

=  rsinm-2xcosna;(ia;—  J  sinTO  x  cosn  a?  da;. 


SEDUCTION  FORMULAS  187 

Substitute,  transpose,  add,  and  divide  by  — ^—  • 

n  -j- 1 

.-.   Csmmxcosnxdx=-smm~lxCOsn+lx  +  ^^  fsinM-2zcoswcedce. 
*/  m-\-n  m  +  nJ 

To  derive  (5) : 

Substitute  m  —  2  =  —  m',  or  m  =  2  —  m',  in  (3),  transpose, 

divide  by ,  and  omit  the  accent. 

n  —  m'  +  2 

The  derivations  of  the  others  are  left  as  exercises  to  the 

student. 

EXERCISES 

Show  by  integration  that : 

,      r    -,       Tan  x     TidM."  x  ,  i 
t&irxdx  =  — h  log  sec  x. 


2.     I  cot3  ccacc  = log  smcc. 


.   9         ,     7        /cosce  ,  cos3 a;     cos5 ce\  since  .    ce 

smi cos  xdx  =  [-8- + ~W— T ) T  + 16 

/~cos4ce$ce  cos3ce       3  3n      ,      x 

4«     I  — —» —  =  — ^   .   „ -cosce  —  -log  tan -• 

J      sm3ce  2sm2ce     2  2     5        2 

/* sin2 x dx _    since         since    _1,      ,      Ar  ,  x\ 
J     cos5  a;        4cos4ce     8cos2ce     8  \I     2  J 

dx  cos  x        4  cos  ce        8      , 

cot  X. 


■f 

:    (— 

J  sin4 


sin6cc         5sin5cc     15  sin3ce     15 

dx        _ 1 5 5  sin  x 

cccos3cc         3  sin3  cc  cos2  cc     3  since  cos2  x     2  cos2  x 

5 

2 


+  glogtan(!  +  !). 


sec6ce  dec  =  tan  ce  f —  +  — ; —  +  — 

ocos4cc     locos^ce     15 


»/ 

9.     I  cosec5ccdcc  =  — cot  cef — ^^t-  +  tz—. ]  +  -logtan-' 

J  \4,sm3x     SsmxJ      8     6        2 


10.     ftan3cesec5ce^  =  ^^-^^. 
J  7  5 


CHAPTER   XIX 


SUMMATION  OP  f(x)Ax.    PLANE  AKEAS  IN  KECTANGULAE 

COORDINATES 

160.    Let  y=f(x)  be  an  equation  in  which  f(x)  is  single 
valued  and  continuous  for  all  values  of  x  between  and  includ- 


ing two  values  a  and  b.      Let 
positive  integer. 


a 


n 


=  Ax,  where  n  is  any 


tr  =  6 


Definition.     The  symbol  2^f(x)Ax  is  used  to  denote 

z  =  a 

f(a)Ax+f(a+Ax)Ax+f(a+2Ax)AxJ \-f(a  +  (n-T)Ax)Ax 

for  any  value  of  n  a  positive  integer. 

x  =  b 

161.    To  interpret  ^,f(x)Ax  geometrically: 

At  first,  suppose  that  f(x)  is  positive  for  all  values  of  x 
between  a  and  b. 

Y 


0 


A      Ax    A2    A3 


Au-i  B       X 


Fig.  56. 


Under  this  supposition  the  curve  y  =f(x)  is  above  the  ic-axis 
for  all  values  of  x  between  a  and  b.  Suppose  that  it  is  as  in 
Fig.  56. 

Let  OA  and  OB  represent  the  abscissas  a  and  b  respectively. 

188 


8  UMMA  TION  OF  f(x)  Ax 


189 


b  —  a 


Divide  AB  into  n  equal  parts.     Each  part  is  of  length 

lb 

Call  each  part  Ace.  Denote  the  successive  points  of  division 
of  AB  by  A,,  A2}  A3,  •  •,  A-i-  At  A  Ai,  A2,  A3,  •••,  An_lf  B, 
erect  ordinates  to  meet  the  curve  in  C,  Bx,  B2,  B3,  •  ••,  Bn_x,  D 
respectively.  From  C,  Bh  B2,  —,  Bn_x,  draw  lines  parallel  to 
the  #-axis  to  meet  the  next  succeeding  ordinates  in  C\,  C2,  C3, 
•  ••,  Cn  respectively. 

1  =  6 

The  symbol  ^)f(x)Ax  denotes  the  sum  of  the  areas  of  the 

x  =  a 

rectangles,  CA1}  BiA2,  B2A3,  •  ••,  JS^-B. 

Next,  suppose  that  f(x)  is  negative  for  all  values  of  x 
between  a  and  b. 

Under  this  supposition,  the  curve  y  =  f(x)  is  below  the 
a>axis  for  all  values  of  x  between  a  and  b.  Suppose  that  it 
is  as  in  Fig.  57. 


Fig.  57. 


Make  the  same  construction  as  in  Fig.  56. 

Since  f(x)  is  negative  for  all  values  of  x  between  a  and  b, 


x  =  b 


x  =  b 


each  term  of  2^f(x)Ax  is  negative.    Then  —  ^)f(x)Ax  denotes 

x  =  a  x=a 

the  sum  of  the  areas  of  the  rectangles  CA1}  BrA2,  B2A3,  •••, 

x  =  b 

B^jB,  and  therefore  j£f(x)Ax  denotes  the  negative  of  the 
sum  of  the  areas  of  these  rectangles. 


190       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Next,  suppose  that  f(x)  changes  in  sign  as  x  increases  from 
a  to  b. 

Under  this  supposition,  the  curve  y=f(x)  crosses  the  a>axis 
for  some  value  or  values  of  x  between  a  and  b. 

Suppose  that  it  is  as  in  Fig.  58. 


A»-xB     X 


Make  the  same  construction  as  in  Fig.  56  or  Fig.  57. 

From  the  preceding  investigation,  it  is  evident  that  in  this 

a  =  b 

case,2/(x)A#  denotes  the  sum  of  the  areas  of  certain  rectan- 

x  =  a 

gles,  minus  the  sum  of  the  areas  of  others.  Thus,  in  Fig.  58, 
it  denotes  the  sum  of  the  areas  of  the  rectangles  from  A  to 
G,  minus  the  sum  of  the  areas  of  the  rectangles  from  G  to  H, 
plus  the  sum  of  the  areas  of  the  rectangles  from  H  to  B,  where 
G  and  H  are  the  first  and  last  points  of  division,  respectively, 
of  the  line  AB,  when  the  curve  is  below  the  a?-axis. 

162.  Definitions.  Those  of  the,  rectangles  CA1}  BXA2,  B2A3,  •  •  •_, 
Bn_xB  of  the  preceding  article  whose  bases  not  on  the  #-axis  lie 
wholly  between  the  curve  and  the  #-axis  are  called  inscribed 
rectangles. 

Thus,  04„  BjA*  B2A3  of  Fig.  56,  or  OAlf  BXA2,  B2A3  of 
Fig.  57,  are  inscribed  rectangles. 

Those  of  the  rectangles  CAlt  BXA2,  B2A3,  •••,  Bn_1B  of  the  pre- 
ceding article  such  that  the  curve  lies  wholly  between  the  base 
not  on  the  cc-axis  and  the  ic-axis  are  called  circumscribed  rectangles. 

Thus,  CA1}  BXA2,  B2A3  of  Fig.  58  are  circumscribed  rectangles. 


PLANE  AREAS,  RECTANGULAR   COORDINATES     191 


EXERCISES 


In  each  of  the  following  problems  plot  the  curve,  draw  the 
rectangles,  and  calculate : 

1.     X  x2  Ax,  (a)  when  n  =  5 ;  (6)  when  w  =  10. 

,4*0.  (a)  2.04 ;  (b)  2.19. 


st=l 
x=2 


2.  ]V  sin  #  Aaj,  when  n  =  4. 

x=0 
x=2 

3.  X  ^°oio  *  &x>  wnen  »  ==  5. 


J.71S.    1.16. 


Ans.  0.137. 


In  Chapter  XYI  we  regarded  area  as  the  inverse  of  a  differ- 
ential. It  is  more  convenient,  however,  to  regard  it  as  the 
limit  of  the  sum  of  a  set  of  rectangles  described  as  explained 
in  Art.  161. 

163.  Let  y  =f(x)  be  an  equation  in  which  f(x)  is  single  valued 
and  continuous  for  all  values  of  x  between  and  including  two 
values  a  and  b.  To  investigate  the  area  between  the  curve 
y=f(x),  the  cc-axis,  and  the  ordinates  corresponding  to  the 
abscissas  a  and  b  respectively,  from  the  point  of  view  of  the 
limit  of  the  sum  of  the  areas  of  a  set  of  rectangles. 

164.  At  first  suppose  that  f(x)  is  positive  for  all  values  of  x 
between  a  and  b. 

Suppose  that  the  curve  is  as  in  Fig.  59.  Let  OA  and  OB 
represent  the  abscissas  a  and  b  respectively.     Divide  AB  into 


0 


A      A;    A2    A3 


At   A 


fc+i 


A»iB 


Fig.  59. 


192       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

n  equal  parts.  Call  each  part  Ax.  Denote  the  successive  points 
of  division  of  AB  by  AY,  A2,  A»,  •  ••,  An_x  respectively.  At  A1} 
A2,  A3,  •  ••,  An_x,  erect  ordinates  to  the  curve.  Let  x^  x2,  x3,  •••, 
xM_i,  denote  the  abscissas  0Aly  OA2,  OA3,  •  •  -,  OAn_Y  respectively. 
The  ordinates  drawn  at  Alt  A2,  AS}  •••,  An_x,  divide  the  area 
up  into  infinitesimal  strips  of  area.  Let  AkA  represent  the 
strip  whose  base  is  AkAk+1.     Then,  by  Art.  141, 


limit 
Ax=0 


'AkA 


=/(«*)• 


Ax 
Therefore,  from  the  definition  of  a  limit, 

Ax 

where  ek  is  infinitesimal  as  Ax  =  0,  or  when  n  =  ao. 

.-.  AkA=f(xk)Ax-\-ekAx. 

Let  7c  take  in  succession  the  values  1,  2,  3,  •••,  n  —  1.     Let  AkA 
and  ek  when  x  =  a  be  denoted  by  A0A  and  e0  respectively. 

Then  A0A=f(a)  Ax-f  e0Ax, 

AXA  —/(x^)  Ax  +  exAx, 
A2^L  =f(x2)  Ax  +  e2Ax, 


An_i^l  =/<X_i)  Ax  +  e^Ax. 

Therefore  A  =  A0^.  +  A^  +  A2A  -\ f-  An_xA 

=/(a)  Ax  4-/(«i)  Ax  +/(x2)  Ax  H h/(a„_i)  Ax 

+  (c0  +  ci  +  e2  H h  e»-i)  Ax 

=/(a)  Ax  +/(a  +  Ax)  Ax  +/(a  +  2  Ax)  Ax  +  •  •  • 
+/(a  +  (w  —  1)  Ax)  Ax 
4.(eo4-Cl  +  e2+.-.  +  ew_1)Aaj 
=  ^/(x)Ax+(e0  +  e1  +  e2+'-  +  en_1)Ax. 


PLANE  AREAS,   RECTANGULAR   COORDINATES     193 


To  show  that 


limit 


(eo  +  q+e.H h  €„..!>  Aa; 


=  0 


\(e0  +  el  +  c>-\ he»-i)|  ^  | Co|  +  |«i|  +  |e2|  H h  |e»-i|- 

Of  the  infinitesimals  |e0|,  |  ex |,  |c2|,  •••,  |en_i|,  let  |cf|  be  one 
which  is  not  less  than  any  of  the  others. 

.-.   |  (e0  +  Cl  +  e2  H h  £„_!)  |  <; nfa]. 

.'.    |(e0  +  ei  +  e2H \- en_x)  Ax\^n\et\Ax. 

Now  n Aa?  =b  —  a. 

.'.    \(eo  +  e1  +  e2-\ r-e„_i)Aa;|  ^  (6  —  a)|cf|. 

When  7i  =  oo,  e*  =  0  and  therefore  (&  —  a)  | et|  =  0. 


limit 
w=oo 

limit 

n=cc 


|(«b  +  €i  +  e2H heM_i)Aaj|    =0. 


165.   Next,  suppose  that  /(#),  in  Art.  163,  is  negative  for 
all  values  of  x  between  a  and  b. 

Suppose  that  the  curve  is  as  in  Fig.  60. 


Fig.  60. 


Make  the  same  construction  as  in  Fig.  59,  Art.  164. 

Let  x1}  x2,  x3,  •••,  xn_1)  denote  the  abscissas  OA1}  OA2,  OA5, 

•,  OAn_x  respectively. 


194       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 
Then,  as  in  Art.  143, 


limit 
Az  =  0 


'AkA 


Ax 
.'.   AkA  =  -f(xh)Ax  +  ekAx. 

By  reasoning  exactly  similar  to  that  employed  in  Art.  164, 
we  find  in  this  case  that : 

x  =  a 

x  =  b 

166.   To  evaluate  (a)  ]™'£  ^f(x)Ax  of  Art.  164,  and  (6) 


(a)  By  Art.  141,  the  area  inclosed  by  the  curve  y=f(x),  the 
#-axis,  and  the  ordinates  corresponding  to  the  abscissas  a  and 

b  respectively  is  <£  (b)  —  <j>  (a),  where    I  f(x)  dx  =  ^>  (x)  -+-  c.    By 
the  preceding  article,  this  area  also  =  ?™^  ^f(x)Ax.     There- 

x  =  a 
x  =  h  _ 

fore  ^™^  ^  f(x) &x  =  4>(P)-4>  (a),  where  J  f(x)  dx  =<\>(x)  +  c. 

x  =  a 

(b)  By  Art.  143,  the  area  inclosed  by  the  cnrve  y  =  f(x) ,  the 
ce-axis,  and  the  ordinates  corresponding  to  the  abscissas  a  and 

b  respectively  is  —  \_^{b)  —  <£(«)],  where  I  f(x) dx  —  <f> (x)  -f  c. 
By  the  preceding  article,  this  area  also  =  —  J^  ^/(£)Aa\ 


Therefore  -^  2)/(«)Aaj  =  -  [<£(&)  -  <K«)]>  where  f/(a>)  c?x 
=  <£(#)  + c. 

167.  As  illustrations  of  the  method  of  application  of  the 
above  principles  to  problems  in  area,  consider  the  following 
examples : 


PLANE  AREAS,   RECTANGULAR   COORDINATES     195 


Example  1.  Find  the  area  inclosed  by  the  curve  4=y 
=  x3  —  6  x2  -+- 11  x  +  4,  the  a>axis,  and  the  ordinates  correspond- 
ing to  the  abscissas  0  and  4  respectively. 

The  curve  is  as  in  Fig.  61. 

The  required  area 

x  =  4 


=  ^2)i(^-6a?l  +  ll»  +  4)A0 


x  =  0 


l(j-™  +  1Jf  +  ** 


x  =  4 


±(t-2x3  +  ^^  +  ±x 
4V4  2 


z=0 


Fig.  61. 


=  10. 


Example  2.  Find  the  area  inclosed  by  the  curve  4y 
=  x3  —  6  x2-\-llx  — 12,  the  #-axis,  and  the  ordinates  corre- 
sponding to  the  abscissas  0  and  4  respectively. 

The  curve  is  as  in  Fig.  62. 

The  required  area 


x  =  4 


x  —  0 


12  x 

Jx- 

12  x 


x 


J-  I X  r»     <?    i    J--L  XT 


x  =  4 


2xr  H — 


Fig.  62. 


=  6. 


4V4 


x  =  0 


168.  Next,  suppose  that  f(x)  in  Art.  163  changes  sign  as  x 
increases  from  a  to  b.  The  curve  y=f(x)  therefore  crosses 
the  x-axis  for  some  value  or  values  of  x  between  a  and  b. 
Since  there  are  different  expressions  for  area  according  as  the 
curve  is  above  or  below  the  a>axis,  we  must,  in  this  case,  find 
the  area  for  each  of  the  regions  above  the  x-axis,  and  for  each 
below,  and  add  the  results. 


196       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


169.  Example.  Find  the  area  inclosed  by  the  curve  4  y 
=  xs  —  6  x2  +  11  #  —  6,  the  #-axis,  and  the  ordinates  correspond- 
ing to  the  abscissas  0  and  4  respectively. 

The  curve  is  as  in  Fig.  63. 

Denote  the  points  where  it  crosses  the  x-axis  by  A,  C,  and 
D.     Let  B  be  the  foot  of  the  ordinate  whose  abscissa  is  4. 

Divide  OA,  AG,  CD,  and  DB  into  %,  n2}  n3,  and  w4  equal 
parts  respectively.  Denote  these  equal  parts  by  A^',  A2£,  A3a;, 
A4x  respectively. 


The  required  area 


limit 


x  =  l 


x  =  0 

+  l^l  X  K*3  -  6^  + 11  x  -  6)  Asx 

x  =  l 

z  =  3 


3  =  2 

x  =  4 


+  ^2i(^3-6^2  +  11^-6)A^ 


Fig.  63. 


2  =  3 


—  U9     I    1    I    I  J.  9\-  6. 

—  4\4'4^4^4>'  —  4* 


2  =  0 

170.   We  have  already  seen  that  ^™^  2J/0*0Aa;  denotes  an 

x  =  a 

area  or  the  negative  of  an  area  according  as  f(x)  is  positive  or 
negative  for  all  values  of  x  between  a  and  b.  This  raises  the 
question  as  to  what  it  would  denote  if  f(x)  changed  in  sign 
as  x  increases  from  a  to  b.     In  Art.  161,  we  saw  that,  in  the 


2  =  6 


particular  case  considered  there,  ^.f{x)Ax  denotes  the  sum 

x  =  a 

of  the  areas  of  the  rectangles  from  A  to  G,  minus  the  sum  of 
the  areas  of  the  rectangles  from  G  to  H,  plus  the  sum  of  the 
areas  of  the  rectangles  from  H  to  B.  (Fig.  58.)  It  can  be 
shown  without  much  difficulty,  although  the  proof  will  not  be 


2=5 


given  here,  that  h^  2)/(a?)Aa;  denotes  the  sum  of  the  areas 


PLANE  AREAS,   llECTANGULAIi   COORDINATES     197 

of  the  regions  ACK  and  MDB,  minus  the  area  of  the  region 

x  =  b 

KFM.     In  general,    ^m**  2)/0»)Aa;  denotes  the  sum  of   the 

x  =  a 

areas  of  all  the  regions  above  the  cc-axis,  minus  the  sum  of 
the  areas  of  all  the  regions  below. 

r  =  b 

171.  To  evaluate     _j£  ^.f(x)Ax  when  f(x)  changes  in  sign 

x  =  a 

as  x  increases  from  a  to  b. 

By  Art.  147,  the  sum  of  the  areas  of  all  the  regions  above 
the  a>axis,  inclosed  by  the  curve,  the  #-axis,  and  the  ordinates 
corresponding  to  the  abscissas  a  and  b  respectively,  minus  the 
sum  of  the  areas  of  all  the  regions  below,  =  <£  (6)  —  <f>  (a),  where 

lf(x)dx  =  <f>(x)  -fa     By  the  preceding  article,  this  difference 

x  =  b  i=& 

also  =  JJ™£  X  f(x) A*«   Therefore  ^  X /(*)**= <tfb)  - <f>(d), 

x=a  x=a 

where  |  f(x)  clx  =  <f>  (x)  +  c,  if  f(x)  is  single  valued  and  continu- 
ous for  all  values  of  x  between  and  including  a  and  b,  and 
changes  in  sign  as  x  increases  from  a  to  b. 

172.  From  Arts.  166  and  171,  we  therefore  see  that 

»=£§/(*)  A*  =  *(&)  -+(«)> 

where  |  /(a?)  dx=  cf>  (x)  +  c,  if  /(a?)  is  single  valued  and  continu- 
ous for  all  values  of  a;  between  and  including  a  and  6,  whether 
it  changes  in  sign  as  x  increases  from  a  to  b  or  not. 


x  =  b 


173.   Definitions.     J1"11^  Xf(x)Ax  'ls  written   {  f(x)dx  and 

read,  "  the  definite  integral  of  f(x)  dx  between  a  and  b." 

The  values  a  and  b  are  called  the  lower  and  upper  limits  of 

integration  respectively. 

The  expression  f(x)  is  called  the  integrand. 


198       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

174.  In  the  last  chapter,  we  saw  that  an  indefinite  integral 
could  sometimes  be  evaluated  by  the  substitution  of  a  new 
variable.  In  a  definite  integral,  if  we  substitute  a  new  vari- 
able, we  must  either  return  to  the  old  variable  before  substi- 
tuting the  limits  in  the  result,  or  else  change  the  limits  at  the 
time  of  substitution  to  suit  the  new  variable. 

Tor    example,   suppose    that    it    is    required    to    evaluate 


x 


(      /     2.  2. 

■\a3  —  x3  dx. 


o 

1  _2 

Let  x3  =  u.         .-.  J  x  3  ax  =  du. 

.-.  dx  =  3  u2du. 


Substitute  in  the  integral. 
"We  may  proceed  as  follows : 


x 


a      I     2  2 

\a3  —  x3dx 


u2\a3  —  u2du 


=3X' 

=  3  [  - 1 V(af  -  u*)*  +jMa3  -  u*  +  a3  sin"1^ 
=  3  {  -  i  V(a*  -  **)'+  v  f**^1  -  **  +  af  sin_1^  i 


i  =  0 


4 


=  T%7ra^ 


Or,  we  may  proceed  as  follows : 

Since  x3  =  u,  .-.  u  =  0  when  a?  =  0,  and  w  =  a3  when  a?  =  a. 
And,  as  x  increases  from  0  to  a,  u  increases  from  0  to  a3. 


x 


%/      2  2 

\a3  — 


x3dx 
=  3  P *  u2^a3-u2du 

=  3  J  —  -  \(a3  —  w2)°  +  -^-[  u\a3  —  u2+  a3  sin  *—  » 


a* 
0 


A^ 


=  yWa; 


PLANE  AREAS,    RECTANGULAR   COORDINATES      199 

Of  these  two  methods  of  procedure  the  latter  is  preferable, 
because  the  work  of  transforming  back  again  to  the  old  variable 
is  usually  more  difficult  than  that  of  transforming  the  limits 
to  suit  the  new  variable. 

EXERCISES 

1.  An  ellipse  whose  semi-major  and  semi-minor  axes  are 
5  feet  and  3  feet  respectively  is  cut  into  two  parts  by  a  line 
perpendicular  to  the  major  axis,  2  feet  from  the  center.  Find 
the  area  of  each  of  the  two  segments  of  the  ellipse. 

Ans.  35.23  sq.  ft. ;  11.89  sq.  ft. 

2.  A  circle  with  its  center  at  the  origin,  and  a  parabola  with 
its  vertex  at  the  origin  and  axis  on  the  a>axis  both  pass  through 
the  point  (2,  2  a/2).     Find  the  areas  inclosed  by  the  curves. 

Ans.  24.35;  13.35. 

3.  A  circle  has  a  radius  of  4  feet.  A  parabola  has  its  ver- 
tex at  the  center  of  the  circle,  and  the  distance  of  its  focus  from 
the  vertex  is  1^-  feet.  Find  the  smaller  area  inclosed  by  the 
two  curves.  Ans.  19.07  sq.  ft. 

In  each,  of  the  five  following  curves,  find  the  area  inclosed 
by  the  curve,  the  #-axis,  and  the  ordinates  corresponding  to  the 
abscissas  set  opposite  the  equation  : 

4.  y  =  sin2  x.  x  =  3,  x  =  4.  Ans.  0.183. 

5.  ^  +  t  =  l.  x  =  3x  =  q.  Ans.  7. 37. 
36      4                                            ' 

6.  -^  +  ^  =  1.  X  =  0}X  =  2.  Ans.  11.48. 
16      9 

7.  y  =  x(x-\-l)(x-\-2).  x  =  —  3,  x  =  S.  Ans.  59. 

8.  y=A_(x-l)(X-3)(x-5).  x  =  -2,  x  =  7.      Ans.  12.854. 

9.  y  =  log10  x.  x  =  \,  x  =  2.  Ans.  0.234. 
10.   Find  the  area  of  the  segment  of  the  circle  x2-\-y2  =  a2 

cut  off  by  the  line  x  =  -  •  Ans.  0.614  a2. 

J  2 


200       DIFFERENTIAL  AN  I)  INTEGRAL   CALCULUS 

11.    Find  the  area  inclosed  between  the  parabola  x2  =  ^ay 
and  the  witch  y  =  — — -  •  Ans.  (2  ?r  —  4)  a2. 

Evaluate  the  five  following  integrals,  changing  the  limits  as 
you  change  the  variable  : 

2 

12  r5(x-3ydx  m        Let(a._3)i  =  M<  Ans.  0.37. 

13  r2  Ve*  dxm  Jjetex  =  u2.  Ans.  2.44 


dx     ■  LetVa;  =  w.  -4?is.  -3.386.. 

15.  L        I       T-  Leta6  =  M.  J-fts.  5.31. 

c/1   2sc2  +  a3 

M 

16.  I     sin  2  a;  da?.  Let2cc  =  i^.  Ans.  0.765. 

17.  Evaluate   I   cos  (2  05—  1)  dx.     Find  the  area  inclosed  by 

the  curve  y  =  cos  (2x  —  1),  the  #-axis,  and  the  ordinates  corre- 
sponding to  the  abscissas  \  and  2  respectively. 

Ans.  0.071 ;  0.929. 

Find  the  area  of  the  loop  in  each  of  the  three  following 
curves : 

18.  xy2  —  (x  —  a)2  (2  a  —  x).  Ans.  i(4^7r)a2. 

19 .  9  y2  =  (as  +  7)  (x  +  4)2.  ^Lws.  f  V3. 

20.  a2y2  =  x\a-x)(2a-x).     Ans.\\±-yj2-%\og  (V2  +  1)|gt. 

21.  Find  the  larger  area  inclosed  by  the  circle  ar2  +  ?/2  =  12  x 
and  the  parabola  y2  =  6x.  Ans.  18  7r  -f-  48. 

22.  Find  tha  area  cut  off  from  the  curve  27  ay2  =  4  (a;  —  2  a)3 
by  the  line  x  =  3a.  ^n§      8  a2 

'  15  V3* 

23.  Find  the  area  of  one  arch  of  the  cycloid  x  =  a(6—  sin  6), 
y  =  a(l  —  cos  0).  -4ws.  3  ira2. 


PLANE  AREAS,    RECTANGULAR  COORDINATES     201 


24.  Find  the  area  inclosed  between  the  y-axis  and  the  arc  of 
the  involute  of  the  circle  ar-f-?/2  =  «2  which  is  traced  as   <£ 

varies  from  0  to   ^.       The    equations    of    the    involute    are 
x = a  (cos  <£  +  (£  sin  <£),?/  =  a  (sin  (f>  —  <j>  cos  <£).      a       no?  fit*  .  .A 

175.  We  have  supposed  thus  far  in  this  chapter  that  fix)  in 
the  equation  y=fix)  is  single  valued  and  continuous  for  all 
values  of  x  between  and  including  a  and  b.  We  shall  now 
suppose  that  fix)  is  single  valued  for  all  values  of  x  between 
a  and  b,  but  violates  the  above  condition  of  continuity  by 
becoming  infinite  or  infinite  negatively  either  for  x  =  a  or  x  =  b 
or  some  value  or  values  of  x  between  a  and  b,  or  by  any  com- 
bination of  these. 

Under  these  suppositions,  we  shall  consider  but  one  case, 
that  in  which  fix)  is  single  valued  and  continuous  for  all 
values  of  x  between  a  and  b,  is  continuous  when  x  =  a,  and 
becomes  infinite  as  x  approaches  b  being  always  less  than 
b.  From  the  reasoning  in  this 
case,  the  student  can  readily 
supply  the  reasoning  in  any  of 
the  other  cases. 

In  the  case  just  mentioned,  the 
curve  may  be  as  in  Fig.  64. 

Let  OA  and  OB  represent  the 
abscissas  a  and  b  respectively. 
On  OB  lay  off  a  distance  OG 
such  that  GB  is  equal  to  some 
arbitrary  length  e.  Then  OG  is  of  length  b  —  e. 
ordinate  GH. 

limit 


Fig.  64. 


Draw  the 


The  area  AGHC  =^   ^  fix)  Ax 


fix)  dx. 


202       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


Suppose  that  I  f(x)  dx  =  <f>(x)  +  c. 

Then  f    ef(x)dx  =  cf>(b  -  e)-  <f>(a). 

If  <J>(b  —  e)  —  <ji(a)  approaches  a  limit  as  e  approaches  zero,  this 
limit  is  called  the  area  between  the  curve,  the  a>axis,  the  ordi- 
nate AC,  and  the  asymptote  x=b.  If  cf>(b  —  e)  —  <j>(a)  increases 
without  limit  as  e  approaches  zero,  the  area  between  the  curve, 
the  a>axis,  the  ordinate  AC,  and  the  asymptote  x  =  b,  is  said 
to  be  infinite. 

176.     Example  1.   Find  the   area    inclosed  by  the   curve 

y  =  —  the   £-axis,   the   ordinate   corresponding   to    the 

-\x  —  1 
abscissa  —  1,  and  the  asymptote  x  =  l. 

The  curve  is  as  in  Fig.  65. 
Let  OA  and  OB  represent  the 
abscissas    —  1    and    + 1    respec- 
tively. 

Let  GB  =  e. 
The  area  AGHC 

>1~e     dx 


-r. 


'X 


Fig.  65. 


=  -  t(*-i)»  1-7 

=  _3£«+S2l. 


2  2  r-\^- 

As  e  approaches  zero,  —  f  e*  + § 2^  approaches  f  23. 
Therefore  the  required  area  =  f  2*. 

Example  2.    Find  the  area  inclosed  by  the  curve  y= — , 

{x     1) 

the  x-axis,  the  ordinate  corresponding  to  the  abscissa  —  1,  and 
the  asymptote  x  ==  1. 

Let  OA  and  OB  represent  the  abscissas  - 1  and  + 1  respec- 
tively (Fig.  66). 


PLANE  AREAS,   RECTANGULAR   COORDINATES     203 


The  area  AGHG 


-jc 


dx 


(x-iy 


l-€ 


Fig.  66. 


x=b- 


2(a;-l)J 

JL  _1 

2  e2     8* 

As   e   approaches    zero,    — -  —  - 
becomes  infinite. 

177.   In  Fig.  64,  as  e  approaches  zero,  A  f(x)  ^x  approaches 
the  sum  of  the  rectangles  inscribed  on  an  equal  number  of 

x—b~ e 

equal   divisions  of  the   line  AB.     Then   if   AaSoS  fix)  ^x 

x=a 

approaches  a  limit  as  e  approaches  zero,  this  limit  can  also  be 

x—b  , 

denoted  by  &™±q%  f(x)^x,  or  J  f(x)dx. 


EXERCISES 

Evaluate  the  following  integrals : 
Xs  dx 


■■■  i 


x- 


xAdx 


Jo  Vl-a;2 

4.      f    ^  .- 
Jo  1  —  s" 

5. 


;-X 


XT 

dx 


a2  +  6V 


xe~ax  dx. 


Ans. 


Ans.   —• 
16 


Ans.  1. 


.4ns.    Infinite. 


-4ns. 


7T 


2a&  . 


a2 


204       DIFFERENTIAL   AND  INTEGRAL    CALCULUS 


7.    Find  the  area  inclosed  by  the  cnrve  ys(x2  —  oFf  =  8  x2, 

2 


f 
2 

the  a>axis,  and  the  asymptote  x  =  a.  Ans.  3  a3" 

8.    Find  the  area  inclosed  by  the  cnrve  y3(x2— a2)2=Sx3,  the 


ic-axis,  and  the  ordinate  whose  abscissa  is  3  a.  Ans.  9  a3. 

9.    Find  the  area  inclosed  by  the  curve  y(x  —  a)2  =  l,  the 
x-axis,  and  the  ordinate  whose  abscissa  is  2  a.       Ans.  Infinite. 

8  a3 

10.  Find  the  entire  area  inclosed  by  the  witch  y  =  — — - — - 

J  *      x2  +  4a2 

and  the  a>axis.  .         .      „ 

Ans.  4  -n-a. 

x3 

11.  Find  the  area  inclosed  by  the  cissoid  y2  =  - ,  the 

n  2a-x 

a>axis,  and  the  asymptote  x  =  2a.  o 

Ans.   -  ttci2. 

2 

12.  Find  the  area  between  the  curve  xy  =  1  and  the  a>axis. 

Ans.  Infinite. 


CHAPTER   XX 


SUMMATION  OP   $\f(P)\**0.    PLANE  AEEAS  IN  POLAE 

COOEDINATES 

178.  Let  r=/(0)  be  an  equation  in  which  /(0)  is  single 

valued  and  continuous  for  all  values  of  0  between  and  includ- 

8  —  a 
ing  two  values   a  and  j3.      Let =  A0,  where  n  is  any 

lb 

positive  integer. 

Definition.     The   symbol    2^J{/(0)}2A0   is   used   to   denote 

lJ/(a)}2A0  +  l-J/(«  +  A0)S2A0  +  iJ/(«  +  2A0)S2A0+- 
+  £{/(«+ (rc-l)A0)}2A0 
for  any  value  of  n  a  positive  integer. 

To  interpret  ^j\f(0)\2^0  geometrically. 

179.  At  first  suppose  that  f{0)  is  positive  for  all  values  of 
6  between  a  and  f3. 

Under  this  supposition, 
the  curve  r  =  /(0)  may 
be  as  in  Fig.  67. 

Let  OB  and  OG  repre- 
sent the  radii  vectores 
whose  angles  are  a  and  ft 
respectively.  Divide  the 
angle  BOO  into  n  equal 
parts.       Each     part     has 

therefore  the  value  — 

n 


Fig.  67. 


205 


206       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Call  each  part  A0.  Denote  the  points  at  which  the  lines  that 
divide  the  angle  BOO  into  n  equal  parts  meet  the  curve  by  Bx, 
B2,  B3,  •••,  Bn_v  With  center  0  and  radii  equal  to  OB,  OBx, 
OB2,  •••,  OBn_x,  describe  arcs  of  circles  to  meet  the  next 
succeeding  radii  vectores  in  C^  C2,  C3,  •••,  Cn  respectively. 

The  symbol  ^)^\f(0)\2A6  denotes  the  sum  of  the  areas  of 
e=a 
the  circular  sectors  OBCl}  OBYC2,  OB2Cs,  •••,  OBn_iCn. 

180.  Next,  suppose  that  f(ff)  is  negative  for  all  values  of  9 
between  a  and  (3,  or  changes  in  sign  as  6  increases  from  a  to  (3. 

Since  f(ff)  is  real,  ^\f(Q)\2  is  positive  for  all  values  of  6 
between  a  and  (3.  Then,  whether  f(0)  is  positive  or  negative, 
the  area  of  each  circular  sector,  described  as  explained  in  the 
preceding  article,  is  iJ/(0){2A0.    Therefore  in  these  cases  also, 

0  =  |3 

2  \\f(P)\2^Q  denotes  the  sum  of  the  areas  of  circular  sectors. 

Q  =  a 

EXERCISES 

In  each  of  the  following  problems,  plot  the  curve,  draw  the 
circular  sectors,  and  calculate : 

!•    2)i0  -cos<9)2A0,  when  n  =  4.  Ans.   0.093. 

0=0 

e=i 
2.    Vsin0A0,  when  n  =  5.  Ans.   0.408. 

0  =  0 

181.  Let  r=f(ff)  be  an  equation  in  which  f(ff)  is  single 
valued  and  continuous  for  all  values  of  0  between  and  includ- 
ing two  values  a  and  ft.  To  find  the  area  inclosed  by  the 
curve  r  =  f(0),  and  the  radii  vectores  that  make  angles  of  a 
and  ft  respectively  with  the  initial  line. 

Suppose  that  the  curve  is  as  in  Fig.  68. 

Let  OB  and  OC  represent  the  radii  vectores  whose  angles 
are  a  and  ft  respectively.     Divide  the  angle  BOC  into  n  equal 


PLANE  AREAS,   POLAR   COORDINATES 


207 


parts.     Call  each  part  A0.     Denote  the  points  at  which   the 

lines    that    divide     the 

angle  BOC  into  n  equal 

parts  meet  the  curve  by 

BlyB2i  J53,  ...,J5„_1.     Let 

the   radii   vectores   OBh 

OB2,     OBs,     ..-,     OBn.l9 

make   the   angles  6lf  82, 

@3)    '")    6n    respectively, 

with  the  initial  line. 

The  radii  vectores  di- 
vide   the    area    up    into    Jf 
infinitesimal     strips     of 
area.     Let  AkA  represent  a  strip  OBk  Bk+l.     Then,  by  Art.  149, 

~A>A~ 


Fig.  68 


limit 

A0  =  O 


AS 


=  «/WF- 


Therefore,  from  the  definition  of  a  limit, 

A  A 

—~  =  ^\f(0k)\2  +  ek  where  ek  is  infinitesimal  as  A0  =  0,  or  when 
Ad 

n  =  oo. 

Let  k  take  in  succession  the  values  1,  2,  3,  •••,  n  —  1.     Let 
AkA  and  eA  when  6  =  a  be  denoted  by  A0A  and  e0  respectively. 

Then  A0A  =  i{J(a)}2A6  +  eoA0, 

A1A=±{f(01)l*Ae  +  e1AO, 

A2A  =  ±\f(ey{2A6  +  e2A0, 


Then,  as  in  Art.  164,  A  =  £?*  X  i\f(6)\2A0. 


e=p 


Therefore  the  required  area  =  ^jj  ^  |-{/(0)52A0. 


208       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


9  =  P 


182.    To  evaluate  ^X  Mffl)\**e' 

6  =  a 

It  was  shown  in  Art.  149  that  the  required  area 

=  008)  -  4>(u),  where  if{f(0)\2d0  =  <j>(0)  +  c. 

Therefore     ^  X  tlf(M*M  =  <l>(p)  -  *(«), 

0=a 


where 


183.  Definitions,     ^jj  $}*{/(0)j2A0  is  denoted  by 

6  =  a 
%J  a 

and  read,  "one  half  the  definite  integral  of  \f(6)\2d$  between 
a  and  p." 

The  values  a  and  f$  are  called  the  lower  and  upper  limits  of 
integration  respectively.  The  expression  \f(6)\2  is  called  the 
integrand. 

184.  Example.  Find  the  area  inclosed  by  the  curve 
r  =  a(l  —  cos  6),  and  the  radii  vectores  that  make  angles  of  0° 
and  180°  respectively  with  the  initial  line. 

The  curve  is  as  in  Fig.  69. 
The  required  area 

=  I™*  XhaX1  ~  cos  ef  ^ 

0  =  0 


=  «2  C\±  _cos  eydo 


Fig.  69. 


=  3-  wa2. 


EXERCISES 


In  each  of  the  three  following  curves,  find  the  area  inclosed 
by  the  curve,  and  the  radii  vectores  that  make  with  the  initial 
line  the  angles  0  set  opposite  the  equation. 


PLANE  AREAS,    POLAR   COORDINATES  209 

1.  r  =  4cos0.     6  =  0,0=1.  Ans.  3.683. 

2.  r  =  6sm$.     0  =  0,0  =  --  Ans.  2.569. 

4 

3.  r  =  4  cos  2  0.     0  =  0,0  =  %.  Ans.  2.909. 

4.  Find  the  area  inclosed  by  one  loop  of  the  curve  r=a  sin  2  0. 

.         ira2 

Ans.  —  • 

5 .  Find  the  area  inclosed  by  the  curve  r  =  a  sin3  -  and  the 
initial  line,  below  the  initial  line.  2 

Ans.  (10tt  +  27V3)^-. 

6.  Find  the  area  inclosed   by  a  small  loop  of   the  curve 
r  =  a  sin  1  0.     Also  the  area  inclosed  by  the  whole  curve. 

Ans.  i(>-2)a2;   i(V  +  2)a2. 

7.  Find    the   area    inclosed    by   the    loop   of    the    curve 
r*cos0  =  a2sm30  *        Sa2     a\     2 

'4        2     8    ' 

8.  Find  the  area  inclosed  by  a  small   loop  of   the  curve 

r2  =  a2  cos  - .  Ans.  ( 2  —  V2)  a2. 

9.  Find  the  area  of  a  loop  of  the  curve  r2  =  a2  cos  2  0. 

Ans.  — 

2 

10.  Find  the  area  inclosed  by  the  curves  r  = and 

1  —  cos  0 
_        4 

+  cos  0  3 

11.  Find  the  area  inclosed  by  the  curve  r=a(sin  2  0  +  cos  2  0). 

Ans.  7ra2. 

12.  Find  the  area  inclosed  by  the  outer  branch  of  the  curve 
r=l+2  sin  J  0  and  the  initial  line,  above  the  initial  line. 

a        7      ,  V3  .  4 
Ans.   -7r+      ~-\ 

6  2      3 

13.  Find  the  area  of  a  loop  of  the  curve  r=2a  sin  3  0. 

Ans.  \ira2. 


CHAPTER   XXI 


THEOKEM  IN  INFINITESIMALS.     DEFINITE  INTEGKAL  IN 

GENEKAL 


x  =  b 


185.   As  already  seen,  each  term  of  £,/(%) &x  of  Chapter  XIX 

x  =  a 

is  the  area  or  the  negative  of  the  area  of  a  rectangle,  two  of 
whose  sides  remain  finite  while  the  other  two  approach  zero 
as  a  limit  as  n  increases  without  limit.     Also,  each  term  of 


0  =  8 


Vi[jf(0)}2A0  of  Chapter  XX  is  the  area  of  a  circular  sector, 

0  =  a 

the  radii  of  which  remain  finite  while  the  angle  between  them 
approaches  zero  as  a  limit  as  n  increases  without  limit.  The 
area,  therefore,  in  either  case  is  expressed  as  the  limit  of  a 
sum  of  terms,  each  of  which  is  infinitesimal  as  the  number  of 
terms  increases  without  limit.  The  student  will  see  later  that 
many  of  the  problems  discussed  in  the  remaining  chapters  of 
the  book  also  involve  the  limit  of  a  sum  of  terms,  each  of 
which  is  infinitesimal  as  the  number  of  terms  increases  with- 
out limit. 

186.  The  limit  of  a  sum  of  terms,  each  of  which  is  infini- 
tesimal as  the  number  of  terms  increases  without  limit,  can 
frequently  be  written  more  simply  by  the  aid  of  the  following 
theorem  in  infinitesimals. 

be  n  infinitesimals,  all  of  the 


Theorem.     Let  ax,  a2,  a3, 


a* 


same  sign,  such*  that   ™^    «i+  «2  +  «3  + 1-  aB  =  -4,  where  A 

is  some  definite  number.     Let  ft,  ft,  ft?  •••,  ft,  be  n  other  infini- 


tesimals such  that 


limit 

11  =  cc 


limit 

11  =  00 


"ft 


ft 


=  1 


limit 

*    n  =  oo 


ft 
do 


|_«n_| 


=  1.    Then  must 


limit 

11  =  cc 


=1, 

/8i+A+ft+ 


limit     P; 


i  «3_ 
fft 


=A. 


THEOREM  IN  INFINITESIMALS 


211 


Proof.     Of  the  ratios   £,  ^2,  £?, 


let  —  be  one  which 


o  «i      «2     «3 

is  not  greater,  and  —  one  which  is  not  less  than  any  of  the 
others.     Then,  by  a  familiar  theorem  of  algebra, 
ft^ft +  ft +  &+-+&.  5;  ft 

«*         «1  +  «2  +  «3  H +  an        «r 

As  ?i  increases  without  limit,  —  and  —  each  approaches  the 

ak  ar 

limit   1.      Therefore,    since    &  +  &  +  &+•••  +  &    is   always 

greater  than  the  one  and  less  than  the  other,  it  also  approaches 
the  limit  1.     And,  by  supposition, 

n=n  L"1  +  «2  +  «3  H h  «WJ  =  -4. 

Therefore     ^  [ft  +  ft  +  ft  +  •  ■  •  +  fi. „]  =  A 

187.   As  an  application  of  this  theorem,  we  shall  again  prove 
that  the  area  defined  in  Art.  163  is  Jj™j£  2)/(«)Aa;. 

Make  the  same  construction  as  in  Fig.  59.      See  Fig.  70. 


Denote  the  abscissas  OAlf  OA2,  OA3, 
xn_x  respectively. 


U-A-n-l)    by  %1)  ®2)  ®3) 


o 


A     Ai  A2    A; 


Fig.  70. 


An-xB      X 


The   required    area  is  the   sum  of  the  strips  whose  bases 
are  AAXi  AXA2,  A2A3,  •••,  An_xB.     Denote  these  strips  by  «1;  a2, 


vu 


a*. 


212       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 
Therefore  the  required  area  =  «x  +  a2  +  as  -\ \-  an 

[«1  +  «2+cj3H !-««],  (1) 


limit 

n=cc 


where  each   strip   is   infinitesimal   as   the   number   of   strips 
increases  without  limit. 

Let  ak+l  denote  any  one  of  these  strips.  Let  x\  and  x"k  be 
the  abscissas  corresponding  to  the  longest  and  shortest  ordi- 
nates  that  can  be  drawn  in  this  strip. 

Then  f(xnk)Ax  <  ak+1  <  f(x'k)Ax. 

Divide  by  f(xk)Ax. 


f(x"k)Ax 


< 


t-k+l 


< 


f(x'k)Ax 


f{xk)Ax       f(xk)Ax      f(xk)Ax 


/(*"*) 


< 


a 


»+i 


< 


/(*'*) 


'/(»*)       /(«*)Aaj      f(xk) 
As  Ax  =  0,  or  when  n  =  oo,  /(a?'*)  and  f(x"k)  both  approach 


.    limit     J\x  k) 


=  1,  and 


limit 


limit 

71  =  00 


a 


k+l 


_f(xk)Ax_ 


71  =  co  [_/(^)  _ 
=  1. 


=  1. 


Therefore,  by  the  theorem  of  the  preceding  article,  ak+1  may 
be  replaced  by  f(xk)Ax  in  any  problem  involving  the  limit  of 
the  sum  of  the  infinitesimals  a.  Let  ak+1  take  in  succession 
the  values  a1?  a2,  a3,  •••,  an,  and  xk  the  values  a,  x1}  x2,  ■•-,  £M_i- 
Substitute  in  (1). 

Therefore  the  required  area  =  j"^  ^/(a)Aa;  +  /(<e1)Aa! 

+  f(x2)Ax-\ h/(a?»-i)Aaj]. 

=  limit  V/(aj)Aa.. 

71  =  oo    -W  •'  v    / 


DEFINITE  INTEGRAL   IN  GENERAL  213 

DEFINITE   INTEGRAL   IN   GENERAL 

x  =  b 

188.    In  ^/(rc)Arc  of  Chapter  XIX,  each  term  is  built  up 

x=a 

from  f(x)  by  multiplying  the  proper  value  of  /(rc)  for  that 
term  by  Arc.     In  Vij/^j^  of  Chapter  XX,  each  term  is 

G  =  a 

built  up  from  /(0)  by  squaring  the  proper  value  of  f(0)  for 
that  term  and  multiplying  the  result  by  \  A0.  The  student  will 
see  later  that  in  all  problems  involving  the  limit  of  the  sum  of 
a  set  of  infinitesimal  terms,  each  term  is  built  up  from  the 
original  function  in  some  way.      Let  F(a)Ax  -f  F(a  +  Arc)Arc 

+  F(a  +  2  Arc)  Ax  -\ f-  F(a  +(n  —  1) Arc)  Arc,  where  Arc  =  h 


n 


be  a  sum  of  terms  built  up  from  the  original  function  in  any 
way.  Suppose  that  F(x)  is  single  valued  and  continuous  for 
all  values  of  rc  between  and  including  a  and  b. 

x  =  b 

189.  Definitions.     The  symbol  V  i^(rc)Arc  is  used  to  denote 

x=  a 

i^(a)Arc-fi^(a  +  Arc)Arc+i^(a+2Arc)ArcH \-F(a+(n— l)Arc)Arc 

for  any  value  of  n  a  positive  integer. 

x  =  b 

n=la  2  F(x)Ax  is   denoted  by  j    F(x)dx,  and  read,   "the 

definite  integral  of  F(x)  clx  between  a  and  b." 

As  in  Art.  173,  a  and  b  are  called  the  lower  and  upper 
limits  of  integration  respectively. 

The  expression  F(x)  is  called  the  integrand. 

190.  To   evaluate    I  F(x)dx,   when  F(x)   is    single  valued 

and  continuous  for  all  values  of  rc  between  and  including  a 
and  b  : 

Plot  the  curve  whose  equation  is  y  =  i^(rc) . 

The  function  F(x)  is  some  function  of  rc,  single  valued  and 
continuous  for  all  values  of  rc  between  and  including  a  and  b. 
The  function  /(rc)  of  Chapter  XIX  is  any  function  of  rc  which 


214       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

satisfies  the  same  conditions.     The  reasoning,  therefore,   ap- 

x  =  6 

plied    in    Chapter    XIX    to    show    that    ^™^  ]T  f(x)Ax,    or 

/-•&  r-  x=a 

I  f(x)  dx,  =  <j>(b)  —  </>(«),  where  I  f(x)dx  =  <j>(x)  +  c,  will  apply 

equally  well  here  to  show  that    I  F(x)dx  =  <f>(b)  —  <f>(a),  where 
F(x)dx=<j>(x)  +  c. 
Therefore,  I  F(x)dx=<f>(b)  —  <f>(a),  where  I  F(x)dx=<f>(x)  +  c. 

EXERCISE 

Prove  by  means  of  the  theorem  of  Art.  186  that  the  area 
defined  in  Art.  181  is  "mfc  ^i5/(0)j2A0. 


CHAPTER   XXII 


SUMMATION   OF  VAar'  +  A/.     LENGTH   OP  AN   AKO   OF  A 

PLANE   CUKVE 

191.  Let  y=f(x)  be  an  equation  in  which  f(x)  is  single 
valued  and  continuous  for  all  values  of  x  between  and  includ- 
ing two  values  a  and  b. 

Suppose  that  the  curve  y  —f(x)  is  as  in  Fig.  71. 

Let  OA  and  OB  represent  the  abscissas  a  and  b  respectively. 


Divide  AB  into  n  equal  parts.  Call  each  part  Ax.  Denote 
the  successive  points  of  division  of  AB  by  Aly  A2,  Ag,  ••-,  An_v 
At  A,  A1}  A2,  As,  ••',  An_i,  B,  erect  ordinates  to  meet  the  curve 
in  C,  B^  B2,  B3,  •••,  Bn_lt  D  respectively.  From  C,  Blf  B2,  •  ••, 
Bn_lf  draw  lines  parallel  to  the  o>axis  to  meet  the  next  suc- 
ceeding ordinates  or  ordinates  produced  in  Clf  C2,  C3,  •  ••,  Cn 
respectively.     Join  CB1}  B^B2,  B2B3,  •••,  Bn_xD. 

Denote  the  abscissas  of  the  points  Alt  A2,  A3,  •••,  An_ly  by 
xx.  x2,  x,  •",  xn_x  respectively.  Denote  the  chords  CBXi  BXB2, 
B2B3,  ••-,  Bn_xD  by  AjC,  A2c,  A3c,  •••,  Anc  respectively.     Then 

215 


216       DIFFERENTIAL  AND  INTEGRAL    CALCULUS 


A1c  =  VA.r  +  AJ/,  where  AJy=f(x1)—f(a), 


A2c  =  VA.v2  +  A2y2,  where  ^=f(x2)  —  f(a$, 


A3c  =  VAar  +  A3y2,  where  A3y=f(x3)—f(x2), 


Anc  =  VAx2  +  Any2,  where  Any  =/(&)  -/(»„_i). 

a:=6 

192.    Definitions.     The   symbol    V  VAa;2  -j-  A?/2  is   used   to 
denote  x=a 


VA^+A^  +  Va^+A^+Va^+A^H h  VAar9+An?/2. 


.7-=?) 


The  symbol  ^  Ac  is  used  to  denote  Axc  +  A2c  -f  A3c  -f h  A„c. 

If  AjS,  A2s,  A3s,  •  •-,  Ans,  denote  the  lengths  of  the  arcs  corre- 
sponding  to    Axc,   A2e,    A3c,   •  ••,    Anc   respectively,   the   symbol 


x  =  6 


Tas  is  used  to  denote  A}s  +  A2s  -f-  A3s  +  •  •  •  +  Ans. 


EXERCISES 

x=3 


1 .    In   the   equation  y  =  se3,  calculate   ^P  Va^2  +  A?/2  when 
n  =  4.  *=2  Ans.  19.03. 


X=7jT 


2.  In  the  equation  y  =  sin  a?,  calculate  ^T  VAsc2  +  A?/2  when 
7i  =  3.  .     *=°  ^4ns.  1.90. 

RECTANGULAR  COORDINATES 

193.  Let  y=f(x)  be  an  equation  in  which  f(x)  is  single 
valued  and  continuous  for  all  values  of  x  between  and  includ- 
ing two  values  a  and  b.  To  find  the  length  of  the  arc  of  the 
curve  y  =f(x)  contained  between  the  two  points  on  the  curve 
whose  abscissas  are  a  and  b  respectively. 

194.  Before  proceeding  to  a  discussion  of  the  problem  stated 
in  the  preceding  article,  we  shall  first  establish  the  following 
theorem  : 


LENGTH   OF  AN  ARC  OF  A   PLANE  CURVE       ZYl 

Theorem.  If  an  arc  of  a  plane  curve  is  concave  toward 
its  chord,  the  limit  of  the  ratio  of  the  arc  to  the  chord  as 
both  approach  zero  is  unity. 

Let  APB  be  an  arc  concave  toward  its  chord  AB  (Fig.  72  or 
Fig.  73).  At  A  aud  B  draw  tangents  to  the  arc  to  meet  in  Q. 
From  Q  draw  QM  perpendicular  to  AB. 


Fig.  72. 


AQ 
AM 


sec  MAQ. 


Fig.  73 


As  AB  approaches  zero,  Z  MAQ  approaches  zero,  and  there- 

AQ- 
AM 


fore    sec  MAQ    approaches    1. 


limit 
AB±0 


limit 
AB=0 


QB' 
MB 


=  1. 


Also, 


=  1.     Therefore,  from  the  fundamental  idea  of  a 


limit,    — —  =  1  +  e,    and    -^ —  =  1  +  e2,    where    e,    and   e2   are 
'    AM  MB 

infinitesimal. 


.-.  AQ  =  AM+AM-ely  and  QB  =  MB  +  MB  ■  e 


*  *2- 


A  Q  +  QB  =  AM  +MB  +  AM.e1  +  MB>ei 


=  AB  +  AM  •  q  +  J^S  •  e2. 


^Q  +  Q5      .    .  AM     .  MB 

7^ =  J-  T 7^7  el  -\ 777 


.45 


.45     *    '     AB 


218       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 
AM 


Since 


limit 
AB±0 


AB 
AM 
AB 


<  1,  and  ex  approaches  zero  as  AB  approaches  zero, 

MB 


—  0       Also     limit 
_U.      A1SO,  AB±Q 


AB 


e., 


=  0. 


limit 
AB±0 


'AQ  +  QB- 

AB 


Now  AB  <  arc  APB  <  A  Q  +  QB. 

Divide  by  AB. 

1      arc  APB     AQ  +  QB 
AB  AB 

Therefore  is  always  greater  than  1  and  less  than  a 

AB  J     5 

number  that  approaches  1. 


Therefore 


limit 
AB=0 


arc  APB' 
AB 


=  1. 


195.    To  return  to  the  problem  of  Art.  193. 

Suppose  that  the  curve  y=f(x)  is  concave  toward  all  chords 
that  can  be  drawn  to  it  from  points  between  the  two  whose 
abscissas  are  a  and  b  respectively,  and  that  the  slope  of  the 
tangent  line  to  the  curve  does  not  become  infinite  at  any  point 
between  and  including  these  two  points. 


0 


A     Ai    A2   A3  ■  Ak  Ajt+i 

Fig.  74. 


A,t-i 


Under  this  supposition  the  curve  may  be  as  in  Fig.  74. 
Make  the  same  construction  as  in  Fig.  71,  Art.  191. 


LENGTH  OF  AN  ABC   OF  A   PLANE  CURVE       219 

x=b 

The  required  length  of  arc  =  V  As 

x=a 

=  limit  V  As 

?i  =  co  jLj 
x~a 

=  limit  yA 


by  the  preceding  article,  since 


limit 


As 

x=b 


=  1, 


=S5;vafta/. 


(1) 


The  chord  BkBk+i  =  VAar'  -f  Ak+1y2.  Compare  this  length 
with  Veto2  +  clk+1y2,  where  dk+1y  is  the  differential  correspond- 
ing to  Ak+1y.  


limit 


_Vdx2  +  dk+1y2_ 


limit 

11  —  CO 


J 


1  + 


cto 


a/ 


1+ 


dk+ly 
dx 


v^c 


dx 


Therefore  Va#2  +  Ak+1y2  may  be  replaced  by  Veto2  +  c4+1?/2 
in  any  problem  involving  the  limit  of  the  sum  of  infinitesimals. 
Replace  each  term  in  (1)  by  Veto2  -f  dk+1y2,  where  dk+ly  has  the 
proper  value  for  that  term. 

Therefore  the  required  length  of  arc 

x=h 


=  n=*X^^+W  (2) 

x—a 


\dxj 


220        DIFFERENTIAL   AND   INTEGRAL    CALCULUS 


by  dividing  and  multiplying  by  dx, 


-jrv 


dx) 


The  length  of  an  arc  can  also  be  expressed  as  j    \/l  +  (  —  )  cfa/ 

by  dividing  and  multiplying  (2)  by  dy.  In  this  case  c  and  d 
are  the  values  of  y  corresponding  respectively  to  the  values 
a  and  b  of  x. 

196.  The  results  of  the  preceding  article  have  been  estab- 
lished only  in  the  case  where  the  curve  satisfies  the  conditions 
imposed  in  that  article.  When  either  or  both  of  these  con- 
ditions are  violated,  a  special  investigation  is  necessary.  It 
would  be  found,  however,  that  in  any  case  in  which  the  equa- 
tion of  the  curve  satisfies  the  conditions  of  Art.  193,  the 
results  are  the  same  as  those  obtained  in  the  case  already  con- 
sidered. As  the  investigation  would  be  tedious,  we  shall 
assume  the  truth  of  this  statement  without  further  discussion. 

197.  Example.  Find  the  length  of  the  arc  of  the  catenary 
y  =  -  (e°  +  e~«)  contained  between  the  two  points  on  the  curve 

whose  abscissas  are  0  and  a  respectively. 
The  curve  is  as  in  Fig.  75. 

Since  y  =  -(e«  +  e  .), -JU= -(<?•- e  -). 

Therefore  the  required  length  of  arc 
Y  limit 

71  =  00 


*=«  x  _x   2 

x=0     ™ 


'1  +  \(ea  —  e  a)  dx 

-J       r*a        x  _x 

=-  I    (e°  +  e  A  dx 


-i" 


LENGTH  OF  AN  ARC  OF  A   PLANE  CURVE       221 


POLAR   COORDINATES 

198.  Let  r=f{6)  be  an  equation  in  which  f(6)  is  single  val- 
ued and  continuous  for  all  values  of  6  between  and  including 
two  values  a  and  ft.  To  find  the  length  of  the  arc  of  the  curve 
r=/(0)  contained  between  the  two  points  on  the  curve  whose 
radii  vectores  make  angles 
of  a  and  ft  respectively  with 
the  initial  line. 

Suppose  that  the  curve 
that  represents  the  equation 
r=/(0)  is  as  in  Fig.  76. 

Let  AOB  and  AOC  rep- 
resent the  angles  a  and  ft 
respectively.  From  C  and 
B  draw  CD  and  BG  perpen- 
dicular to  OA. 

Suppose  that  B  and  O 
have  the  abscissas  a  and  b  respectively  when  the  equation 
of  the  curve  is  expressed  in  rectangular  coordinates.  Then, 
in    rectangular   coordinates,    the    required    length    of    arc    is 

x—b 

limit  >g  yjdrg  _j_  ciy2^     Transform  this  expression  to  polar  coor- 

x=a 

dinates.     The  equations  of  transformation  are 

x  =  r  cos  0, 

y  =  r  sin  6. 
Since  x  =  r  cos  0,     .:  dx  =  cos  6  dr  —  r  sin  0  d$. 
Since  y  =  r  sin  0,     .-.  dy  =  sin  0  dr  +  r  cos  0  dO. 


Fig.  76. 


x=b  x=b 

limit  "N^  -.  /V7™2  i   ,7„, 2 limit 


V  Vdx2  +  df=  ^  ]P  Vdr2  +  rW. 


Now  0  =  a     when  x  =  a,     and  0  =  /?     when  a;  =  6. 


222       DIFFERENTIAL  AND  INTEG11AL    CALCULUS 


Therefore  the  required  length  of  arc  =    _      V  Vc?r2  +  r2d6 


=£ij*  +  ffia 


\ciej 


The  length  of  the  arc  can  also  be  expressed  as 


xv> 


f1  +  r2/^  , 

\clry 


by  dividing  and  multiplying  by  dr.  In  this  case,  y  and  8  are 
the  values  of  r  corresponding  to  the  values  a  and  (3  respectively 
of  6. 

199.    Example.      Find  the  entire  length  of  the  cardioide 
r  =  a(l  —  cos  6). 

The  curve  is  as  in  Fig.  77, 
The  required  length  of  arc 

=  2  f  Va2(l  -  cos  Of  +  dz  sin2  6  d$ 

=  2  a  rV2(l-cos0)  c?0 
«y  o 

=4aX 


sin  -  d$ 


Fig.  77. 


=  8  a. 


EXERCISES 

1.  Find  the  length  of  the  arc  of  the  curve  y  =  log  x  con- 
tained between  the  points  on  the  curve  whose  abscissas  are  1 
and  5  respectively.  Ans.  4.37. 

2  2 

2.  Find  the  length  of  the  four  cusped  hypocycloid  xJ -\-y*  =  l 
contained  between  the  points  on  the  curve  whose  abscissas  are 
—  \  and  1  respectively.  Ans.  2.445. 

3.  Find  the  length  of  the  arc  of  the  four  cusped  hypocycloid 
xi  _j_  yi  —  af  first  using  the  limits  0  to  a,  then  the  limits  —  a 
to  a.  Ans.  6  a. 


LENGTH  OF  AN  ARC  OF  A    PLANE  CURVE        22-\ 

4.  Find  the  length  of  the  loop  of  the  curve 

9  y2  =  (x  +  7)  (x  +  4)2.  Ans.  4V3. 

5.  Find  the  length  of  the  arc  of  the  curve  y  =  log  sec  x  con- 
tained between  the  points  on  the  curve  whose  abscissas  are  0  and 

5  respectively.  Aiis.  log  (2  -f  V3)  =  1.32. 


6.  Find  the  length  of  the  parabola  ?/2  =  4  ax  cut  off  by  the 
latus  rectum.  Ans.  2  [  V2  +  log  (1  +  V2)]  a  =  4.591  a. 

7.  Find  the  length  of  the  arc  of  the  curve  y  =  log  (1  —  x2) 
contained  between  the  points  on  the  curve  whose  abscissas  are 
0  and  x  respectively.  j^ns    jQCT  1  +  x  _ 

1  —  x 

8.  Find  the  length  of  the  arc  of  the  curve  y3  =  ax2  contained 
between  the  point  (0,  0)  and  the  point  (x,  y)  on  the  curve. 

Ans.  (4"  +  9y)*. 

27  a* 

9.  Find  the  length  of  an  arch  of  the  cycloid  x  =  a(6  —  sin  6), 
y  =  a(l  —  cos  &).  Ans.  8  a. 

10.  Find  the  length  of  the  arc  traced  by  the  extremity  of  a 
string  while  it  is  being  unwound  from  a  circle.  Given  that  the 
equations  of  the  involute  of  the  circle  are  x  =  <x(cos  <j>  +  <£  sin  <£), 
y  =  a  (sin  <£  —  <£  cos  <£).  Ans.  2  ?r2a. 

11.  Find  the  length  of  an  arc  of  the  hypocycloid 

x  =  (a  —  b)  cos  <£  +  b  cos  a~    <f>,  y  =  (a  —  b)  sin  <f>  —  b  sin  a         </> 

described  in  a  complete  revolution  of  the  generating  circle. 

Ans.  «6(°-6). 
a 

12.  The  cables  of  a  suspension  bridge  hang  in  the  form  of  a 
parabola.  Given  that  the  length  of  the  bridge  is  1000  feet  and 
that  the  distance  from  the  lowest  point  of  the  cable  to  the  level 
of  the  top  of  the  piers  is  50  feet,  find  the  length  of  a  cable. 

Ans.  1006.4  ft. 


224       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

13.  Find  the  length  of  the  arc  of  the  circle  r  —  2  a  cos  6. 

Ans.  2  -n-a. 

14.  Find  the  length  of  the  arc  of  the  hyperbolic  spiral  rO  =  a 
contained  between  the  points  on  the  curve  where  0  =  ^  anc^ 
0  =  J  respectively.  Ans.  [if  +  log  f ]  a. 

15.  Find  the  length  of  the  arc  of  the  spiral  of  Archimedes 
r  =  aO  contained  between  the  points  on  the  curve  where  6  =  0 
and  0  =  2tt  respectively. 

Ans.  [tt  Vl  +  4  7T2  +  i  log  (2  7T  -f  Vl  +  4  tt2)  ]  a. 

16.  Find  the  length  of  the  arc  of  the  parabola  r  =  - -, 

8  e  1  +  cos  6 

contained  between  the  points  on  the  curve  where  0  =  0  and 


0  =  -  respectively.  Ans.  2 

Li 


V2  +  log  tan  — - 


17.  Find  the  length  of  the  arc  of  the  curve  rF  =  a?  sin  -• 

o 

Ans.  f  ttcl. 

18.  Find  the  length  of  the  arc  of  the  cissoid  r  =  20  tan  6  sin  6 
contained  between  the  cusp  and  the  point  on  the  curve  where 

6  =  --  Ans.  14.6. 


CHAPTER   XXIII 

AKEAS  OP  SUKFAOES  OF  KEVOLUTION 

200.  Let  y  =f(x)  be  an  equation  in  which  f(x)  is  single 
valued  and  continuous  for  all  values  of  x  between  and  includ- 
ing two  values  a  and  b.  To  find  the  area  of  as  much  of  the 
surface  generated  by  the  revolution  of  the  curve  y  =f(x) 
about  the  #-axis  as  is  contained  between  planes  perpendicular 
to  the  a^axis  through  the  points  whose  abscissas  are  a  and  b 
respectively. 

Suppose  that  f(x)  is  positive  for  all  values  of  x  between 
a  and  b. 

Under  this  supposition  the  curve  y=f(x)  may  be  as  in 
Fig.  78. 

Let  OA  and  OB  represent  the  abscissas  a  and  b  respectively. 
Divide  AB  into  n  equal  parts.     Call  each  part  Ax.     Denote 


0 


A     Ai    A2   As 


Afc  A 


ft-H 


An-xB 


Fig.  78. 


the  successive  points  of  division  of  AB  by  Alf  A2,  A3,  -•>,  A^_lm 
At  A,  Alt  A2,  A3,  ••-,  An_lf  B,  erect  ordinates  to  meet  the  curve 
in  C,  B1}  B2,  B3,  •••,  Bn_1}  D  respectively.     Join  CB1}  BXB2, 


B2B3,    ~,  Bn_1D. 


225 


226       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Let  xly  x2,  x3j  •  ••,  xn_lf  denote  the  abscissas  0A1}  0A2,  0A3,  •  ••, 

0An_Y  respectively. 

Since  the  sum  of  the  chords  CB^  BJS^  B2B3,  •  ••,  Bn_xD, 
approaches  the  length  of  the  arc  as  its  limit  as  the  number  of 
chords  increases  without  limit,  we  should  expect  that  the  sum 
of  the  areas  of  the  surfaces  generated  by  the  chords  would 
approach  the  area  of  the  surface  generated  by  the  arc  as  its 
limit  as  the  number  of  chords  increases  without  limit.  On 
the  assumption  that  such  is  the  case,  the  area  of  the  surface 
generated  by  the  arc  can  be  found  as  follows : 

Each  of  the  chords  CBlf  BXB2,  B2B3,  •  ••,  Bn_1D,  generates 
the  frustum  of  a  right  circular  cone.  The  curved  surface  of 
the  frustum  of  a  right  circular  cone  is  the  product  of  the  slant 
height  and  one  half  the  sum  of  the  circumferences  of  the  bases. 
Therefore  the  sum  of  the  areas  of  the  surfaces  generated  by 

the  chords  =  «-[{2/(o)  +  A#}  AlC  +  J2/(0  +  A2y\ A2c 

+  \2f(x2)  +  A32/|A3c  +  -.+  {2/0*W-i)  +  Art2/]Anc]. 
Therefore  the  required  area  of  the  surface 

=  £?*  *-£{2/(a)  +  A#j AlC  +  \2f(x,)  +  A2y\ A2c 
+  \2f{x2)  +  A3y\A3c  +  .-  +  {2/(3^0  +  Any\  AMc]. 

The  area  of  the  surface  generated  by  the  chord  BkBk+l  is 
ir\2f(xk)+Ak+1y\Ak+1c.  (2.)    Compare  this  area  with  2  7rf(xk)Ak+1c. 

Since  Ak+ly  approaches  zero  as  n  increases  without  limit, 
therefore 


limit 
w=co 


•7rf2/(^)+Am?/iA,+1c-|  =  L 
2  7rf(xk)Ak+1C  J 


Therefore  each  term  in  (1),  by  the  theorem  of  Art.  186,  may 
be  replaced  by  2  vf(xk)Ak+1c  where  k  has  the  proper  value  for 
that  term.  Let  k  take  in  succession  the  values  0,  1,  2,  3,  •••,  n. 
Denote  xk  when  k  =  0  by  a. 


AREAS  OF  SURFACES   OF  It  EVOLUTION  227 

Therefore  the  required  area  of  the  surface 

=  i=£  [2  «■-{/(«)  <v  +/(-'-i) A^  +  /0%)*tf  +  •  •  •  +./K-i)Anc }  ] 

2=6 

=  lhnit  Y2tt/Y.x-)Ac 


x  =  b 

limit 

71=00 


£  2  *•/(*)  VAaf'+A*/- 


=2-X>^Vi+(l)^  « 


The  required  area  of  the  surface  cau  also  be  expressed  as 

2  7T 


-jp^Msh 


by  dividing  and  multiplying  by  cfa/.  In  this  integral,  c  and  c? 
are  the  values  of  y,  corresponding  to  the  values  a  and  b  ol  x 
respectively. 

201.  In  the  equation  y—f(x)  of  the  preceding  article,  if  x 
is  a -single- valued  function  of  y  between  the  values  c  and  d  of  y, 
and  is  positive  for  all  values  of  y  between  c  and  d,  the  area  of 
as  much  of  the  surface  generated  by  the  revolution  of  the 
curve  y=f(x)  about  the  2/-axis  as  is  contained  between  planes 
perpendicular  to  the  y-axis  through  the  points  on  the  curve 
whose  abscissas  are  a  and  b  respectively  can  be  found  in  a 
similar  manner  to  be 


7T   I      SCA/1  +(—  ) 


"•r-V1 +(£)'* 


228       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


EXERCISES 

1.  If  f(x)  of  Art.  200  is  negative  for  all  values  of  x  between 
a  and  b,  show  that  the  area  of  the  surface  defined  in  that 

2.  If  x  of  Art.  201  is  negative  for  all  values  of  y  between 
c  and  d,  show  that  the  area  of  the  surface  defined  in  that 
article  is  nh      \        /,  x2 


202.  As  illustrations  of  the  application  of  the  above  prin- 
ciples to  problems  in  areas  of  surfaces,  consider  the  following 
examples  : 

Example  1.  Find  the  area  of  as  much  of  the  surface  gen- 
erated by   the  revolution  about  the  cc-axis   of  the   catenary 

a  f  ^*  a 

y  =  -  I  ea  _|_  e~a  ]  as  is  contained  between  planes  perpendicular 


to  the  cc-axis  through  the  points  whose  abscissas  are  0  and  a 

respectively. 

The  required  area  of  the  surface 


0 


=-j;w^(ij- 


=2"f 


e"  +  e 


dx 


trOf 


Fig.  79.  4    V  e' 

Example  2.     Eind  the  area  of  as  much  of  the  surface  gen- 
erated  by  the  revolution   about  the   ?/-axis  of   the   catenary 

y  =  -(e"  +  c~")  as  is  contained  between  planes  perpendicular 

to  the  ?/-axis  through  the  points  whose  abscissas  are  0  and  a 
respectively. 


AREAS   OF  SURFACES   OF  REVOLUTION 
The  required  area  of  the  surface 


229 


7T  f  x((ia  +  e~a)dx 

XX  XX 

tt \_axea  —  a2ea  —  axe  a  —  a2e  "J", 


=  27ra2  1-- 


by  integration  by  parts 

EXERCISES 

1.  Find  the  area  of  as  much  of  the  surface  generated  by 
the  revolution  about  the  ic-axis  of  the  parabola  y2  =  4  ax  as  is 
contained  between  the  origin  and  the  plane  x  =  a. 

Ans.   f(V8-l)7ra2. 

2.  Find  the  area  of  the  surface  of  the  torus  formed  by  the 
revolution  of  the  curve  x2  -f  (y  —  b)2  =  a2,  b  >  a,  about  the 
X-axis.  .  Ans.   4,ir2ab. 

3.  Find  the  area  of  the  smaller  surface  formed  by  the  revo- 
lution of  the  curve  x2  +  (y  —  b)2  =  a2,  b<  a,  about  the  cc-axis. 

■VaJ-W 


Ans.    4  7r« 


Va2  —  b2—b  sin-1 


a 


4.  Find  the  area  of  the  surface  generated  by  the  revolution 

2  2  2 

of  the  four  cusped  hypocycloid  xJ  +  y*  =  a1  about  the  cc-axis. 

Ans.  ^--n-a2. 

5.  The  axis  of  a  parabolic  reflector  is  20  inches  long  and 
the  focus  is  4|-  inches  from  the  vertex.  Find  the  area  of  the 
surface  of  the  reflector.  Ans.  686  -n-  sq.  in. 

6.  Find  the  area  of  the  surface  generated  by  the  revolution 
of  one  arch  of  the  cycloid  x  =  a(0  —  sin  $),  y  =  a(l  —  cosO) 
about  the  x-axis.  Ans.  -6g4-  ira2. 


CHAPTER  XXIV 


VOLUMES  BY  MEANS   OF  PARALLEL   OEOSS   SECTIONS 


203.  If  the  areas  of  the  cross  sections  of  a  solid  made  by  a 
set  of  parallel  planes,  drawn  in  any  convenient  manner,  can  be 
expressed  in  terms  of  the  distances  of  these  planes  from  a 
fixed  point,  the  volume  of  the  solid  can  readily  be  determined. 
The  following  example  will  illustrate  the  method. 

Example.  Find  the  volume  of  the  right  circular  cone  of 
height  h  and  radius  of  base  a. 

C 

•     Bfc- 


Fig.  81. 


Take  the  vertex  of  the  cone  as  the  origin,  and  the  axis  of 
the  cone  as  the  aj-axis  (see  Fig.  81).  Let  B  denote  the  point 
at  which  the  cc-axis  cuts  the  base  of  the  cone.  Divide  OB  into 
n  equal  parts.*  Call  each  part  As.  Denote  the  successive 
points  of  division  of  OB  by  Alf  A2,  As,  •-..  Through  each 
of  these  points  pass  a  plane  perpendicular  to  the  a-axis.  Let 
the  planes  through  Ak  and  Ak+1  intersect  OC  in  Bk  and  Bk+1 
respectively.      On  the  circles   of   intersection    of   the   planes 

230 


VOLUMES  BY  PARALLEL    CROSS   SECTIONS        231 


through  Ak  and  Ak+1  with  the  cone  construct  right  cylinders 
of  height  Ax. 

Denote  the  abscissas  of  Alf  A2,  A3,  •••  by  xl}  x2,  x3,  •••  respect- 
ively. Denote  the  radii  of  the  circles  of  intersection  of  the 
planes  through  Ak  and  Ak+1  with  the  cone  by  yk  and  yk+l 
respectively. 

From  similar  triangles, 

OAk:AkBk::OB:BO. 

.-.  xk  :yk::h:  a. 


.-•  yk 


h 


Similarly, 


Vk+l  — 


a(xk  +  Ax) 


Let  a1}  a2,  a3,  •••,  an,  denote  the  elements  of  volume  into 
which  the  cone  is  divided  by  the  planes  through  A1}  A2,  A3, 
•  ••,  perpendicular  to  the  a>axis. 

The  volumes  of  the  right  circular  cylinders,  the  radii  of 

2      2  2/  i      a     \9 

whose  bases  are  yk  and  yk+1,  are  ^??L  Ax  and   —  ^_+      ^  Ax 

Ax<ak+1<-- 


respectively. 


K1 


7TCl?Xk2 


7ra2(xk  -f-  Ax)2 
h2 


Divide  by  ^  _„*  Ax. 


h2 


.'•  1< 


«*+i      <  (%k  +  Ax)s 


Tra2x2  Ax 


Xi. 


As  Ax  =  0,  or  when  n  =  ao}  limit 


'(xk  +  Ax) 


2  "I 


&;. 


=  1. 


limit 

71  =  00 


«*+i 


irCTXifAx 
~~h2 


=  1. 


232       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


by 


Therefore,  by  the  theorem  of  Art.  186,  ak+1  may  be  replaced 

ira2X,2Ax 


h2 


in  any  problem  involving  the  limit  of  the  sum  of 


the  infinitesimals  a. 

Now  the  volume  of  the  cone 


limit 

n  =  oo 


Let  ak+l  take  in  succession  the  values  «1?  a2>  «s? 
ic&  the  values  0,  x1}  x2,  •••,  #„_!.     Substitute  in  (1). 
Therefore  the  volume  of  the  cone 


a) 

■,  an,  and 


limit 

n=co 


^lw  +  x12  +  x22  +  ..-+x2n_1\Ax 


limit  yrtT^ 
/iVo 


204.  Of  the  solids,  the  areas  of  cross  sections  of  which  made 
by  a  set  of  parallel  planes  can  be  expressed  in  terms  of  the 
distances  of  these  planes  from  a  fixed  point,  an  important 
class  is  those  formed  by  revolving  a  plane  curve  about  a 
straight  line  in  its  plane. 

Let  y—f(x)  be  an  equation  in  which  f(x)  is  single  valued 
and  continuous  for  all  values  of  x  between  and  including  two 
values  a  and  b.  To  find  the  volume  of  as  much  of  the  solid 
generated  by  revolving  the  curve  y=f(x)  about  the  #-axis  as 
is  contained  between  planes  perpendicular  to  the  a>axis  through 
the  points  whose  abscissas  are  a  and  b  respectively. 

Suppose  that  the  curve  is  as  in  Fig.  82. 

Let  OA  and  OB  represent  the  abscissas  a  and  b  respectively. 
Divide  AB  into  n  equal  parts.     Call  each  part  Ax,     Denote 


VOLUMES  BY  PARALLEL   CROSS   SECTIONS       233 


the  successive  points  of  division  of  AB  by  A1}  A2)  A8,  •••,  An_v 
At  each  of  these  points  draw  the  ordinate  to  the  curve.  Draw 
Bk0k+1  and  Bk+lDk  parallel  to  the  a>axis. 


A      Ai    A2    A3  A&    h/c+i      A»._ib         A 

Fig.  82. 

Denote  the  abscissas  of  A1}  A2,  As,  •••,  An_lf  by  xx  x2  xs  •••,  xn_x 
respectively. 

Let  al}  a2,  a3,  •>>,  an,  denote  the  elements  of  volume  into  which 
the  given  solid  is  divided  by  the  planes  generated  by  the 
revolution  about  the  cc-axis  of  the  ordinates  from  A,  Ah  A2, 

^■39   '"}  -^-n-l)  B- 

.-.  Tr\f(xk)}2kx <  ak+1  <7r\f(xk  +  Ax) }2Ax. 
Divide  by  7r\f(xk)\2\x. 


.-.!< 


a 


k+l 


< 


As  Ax  =  0,  or  when  n  =  00,  limit 


\f(xk  +  Ax)}[ 
f(xkys%x^   -    \f(xk)}* 

f(xk  +  Ax)}*- 


limit 


a 


■k+i 


jif^y^xj 


f/te)}! 


=  1. 


=  1. 


Therefore,  by  the  theorem  of  Art.  186,  ak+1  may  be  replaced 
by  7T  £/(%.)  5  2Aa?  in  any  problem  involving  the  limit  of  the  sum 
of  the  infinitesimals  a. 

Now  the  required  volume  =  [«j  +  «2  +  as  +  •••  +  ««] 

=  ^[«l+«2  +  «3+-+«J.     (1) 


234       BIFFEBENTIAL  AND  INTEGRAL   CALCULUS 

Let  ak+1  take  in  succession  the  values  a1}  a2,  a3,  •••,  an,  and 
xk  the  values  a,  x1}  x2,  xs,  ••-,  xn_v     Substitute  in  (1). 
Therefore  the  required  volume 


+s/(^)j2+-+5/fe-i);2)A^] 

=  limit  vj/(x)j2Aa; 

x=a 

=  7r  I    J/(#)}2cfa,  or 


y2dx. 


205.  In  the  equation  y  =f(x)  of  the  preceding  article,  if  x  is 
a  single-valued  function  of  ?/,  the  volume  of  as  much  of  the 
solid  generated  by  the  revolution  of  the  curve  y=f(x)  about 
the  2/-axis,  as  is  contained  between  planes  perpendicular  to 
the  i/-axis  through  the  points  on  the  curve  whose  ordinates 
are  c  and  d  respectively  can  readily  be  found  to  be 


■  I    x*dy. 


206.  Example  1.  Find  the  volume  of  the  solid  generated 
by  the  revolution  about  the  x-axis  of  as  much  of  the  parabola 
y2  =  2mx  as  is  contained  between  the  origin  and  the  plane 
perpendicular  to  the  x-axis  through  the  point  whose  abscissa 

is  2  m. 

The  required  volume 


r=2m 


limit    V  2irmXdx 

i 

X  CIX 
0 

Fig.  83.  =  4  7rm3. 


/»2« 

=  2  7rm  I 

*/0 


VOLUMES   BY  PARALLEL   CROSS   SECTIONS        235 


Example  2.  Find  the  volume  of  the  solid  generated  by  the 
revolution  about  the  ?y-axis  of  as  much  of  the  parabola  y2=2mx 
as  is  contained  between  the  origin  and  the  plane  perpendicular 
to  the  ?/-axis  through  the  point  whose  abscissa  is  2  m. 

The  required  volume 


4  71l2Jo 


§  7rm3. 


'kh 


yAdy 


EXERCISES 

1.  A  tree  is  24  inches  in  diameter.  A  notch  is  cut  to  the 
center  of  the  tree,  one  face  being  horizontal  and  the  other 
inclined  45°  to  the  horizontal.  Find  the  amount  of  wood  cut 
out.  Ans.  -|  cu.  ft. 

2.  A  variable   equilateral    triangle   moves  with   its   plane 

perpendicular  to  the  x-axis  and  the  ends  of  its  base  on  the  parts 

of  the  curves  y2  =  16ax  and  y2  =  4^ax  respectively  above  the 

ic-axis.     Find  the  volume  generated  by  the  triangle  as  it  moves 

from  the  origin  to  the  points  whose  abscissa  is  a.  V3   3 

Ans.  a  - 

2 

3.  The  cap  of  a  banister  post  is  a  solid,  any  cross  section 
of  which  is  a  square.  A  vertical  plane  through  the  diagonal 
of  the  base  contains  one  diagonal  of  any  such  square,  and  the 
ends,  of  this  diagonal  lie  on  a  circle  of  radius  8  inches  with  its 
center  4  inches  above  the  plane  of  the  base.  Find  the  volume 
of  the  cap.  Ans.  1152  cu.  in. 

4.  Find  the  volume  of  the  torus  formed  by  the  revolution 
of  the  curve  x2  +  (y  —  b)2  =  a2,  b  >a  about  the  x-axis. 

Ans.  27r2a2b. 


236        DIFFERENTIAL  AND  INTEGRAL    CALCULUS 

5.    Find  the  volume  of  the  circular  spindle  formed  by  the 
revolution  about  the  #-axis  of  the  part  of  the  curve  x2-\-(y  —  b)2 


—  r,2 


a'6<c(-  Ans.  2 


~b2  +  2  a2     ,— — to        97  ,  6" 

— — v  cr  —  ¥—  a2b  cos-1  - 

o  a 


6.  Find  the  volume  of  as  much  of  the  solid  generated  by 

x 
the  revolution  of  the  curve  y  =  a  sin-1  -  about  the  a>axis  as  is 

a 

contained  between  the  origin  and  the  plane  perpendicular  to 
the  ie-axis  through  the  point  whose  abscissa  is  a. 

Ans.  ^[y_8]. 

4  J 

7.  A  steel  band  is  placed  on  a  cylindrical  boiler.      A  cross 

section  of  the  band  is  a  semi-ellipse  whose  semi-axes  are  3  and 

3 

— -  inches  respectively,  the  greater  being  parallel  to  the  axis  of 

V6 

the  boiler.     The  diameter  of  the  boiler  is  48  inches.     Find  the 

volume  of  the  band.  Ans.  Q>ir\l  +  6V6?r]  cu.  in. 

8.  Find  the  ratio  between  the  volume  of  the  spindle  formed 
by  the  revolution  of  a  parabola  about  an  ordinate  and  the 
volume  of  the  circumscribing  cylinder.  Ans.  8  :  15. 

9.  Find  the  volume  of  the  solid  formed  by  the  revolution 

of  one  arch  of  the  curve  y  =  2  sin  —  about  the  oj-axis. 

10 

Ans.  20  it2. 

10.  A  bead  is  of  the  form  of  a  sphere  through  which  a 
round  hole  has  been  bored,  the  axis  of  the  hole  coinciding  with 
a  diameter  of  the  sphere.  If  the  diameter  of  the  hole  is  two 
thirds  of  the  radius  of  the  sphere,  compare  the  volume  bored 
out  with  the  volume  of  the  sphere.  Ans.  27  — 16  V2  :  27. 


CHAPTER   XXV 

SUCCESSIVE   INTEGRATION 

207.  In  finding  the  integral  of  a  function  of  two  or  more 
independent  variables,  we  hold  all  the  variables  constant 
except  the  one  with  respect  to  which  we  are  performing  the 
integration,  and  treat  the  function  as  if  it  were  a  function  of 
one  variable  alone. 

For  example,  in  finding  j  x2y2dx,  we  hold  y  constant,  and 
treat  x2y2  as  if  it  were  a  function  of  x  alone. 


Thus,  I  x2y2dx  =  —^~ 


This  is  done  only  when  the  variables  are  independent. 
We  saw  in  finding  the  lengths  of  arcs  of  curves,  in  the  case 
of  two  variables  when  one  was  dependent,  that  we  had  to 
express  one  in  terms  of  the  other,  or  both  in  terms  of  a  third 
variable  before  the  integration  could  be  performed. 

208.    Let  f(x,  y)  be  a  function  of  two  independent  variables. 

Definitions.      J  f(x,  y)dx,  or   I  f(x,  y)dy,  or    I  f(x,  y)dz  where 

z  is  any  third  variable,  is  called  a  single  integral  of  f(x,  y). 

An  integral  of  any  single  integral  of  f(x,  y),  with  respect  to 
any  variable,  is  called  a  double  integral  of  f(x,  y). 

/[   C  1 

\    I  f(x,  y)dx  \  dy  is  a  double  integral  of  f(x,  y). 

An  integral  of  any  double  integral  of  f(x,  y),  with  respect 
to  any  variable,  is  called  a  triple  integral  of  f(x,  y). 

237 


238       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Thus,  I   j    I  f(x,  y)  dx  V  dy   dz   is    a   triple   integral   of 

f(p,  y)- 

These  definitions  are  given  for  definiteness  for  a  function  of 
two  independent  variables.  They  would  be  similar,  however, 
for  a  function  of  one  variable  or  of  any  number  of  independent 
variables. 

The  process  of  taking  integrals  in  succession  of  any  given 
function  is  called  successive  integration. 

209.  The  order  in  which  the  integration  is  to  be  performed 
may  be  indicated  by  brackets  as  in  the  preceding  illustrations, 
or  by  any  other  convenient  notation.  We  shall  employ  the 
notation  most  commonly  adopted  in  text-books,  which  is  to 
write  the  integral  without  brackets,  and  arrange  the  differen- 
tials so  that  their  order,  reading  from  right  to  left,  will  indicate 
the  order  in  which  the  integration  is  to  be  performed. 

Thus,    I  J    j  I  I  dz  )  dy  \  dx  will  be  written    I    I    j  dx  dy  dz. 


EXERCISES 

Show  that : 

1.   £  £  (x2  +  f)  dx  dy  =  I*  (a2  +  62). 
2  •     f~afX  (a2  +  y2)  dx  dy  =  \6-  a4. 
I  r  sin  0d0dr  =  |ol 

4.  II     (x  +  y)dydx  =  ±a5. 

y± 

5.  '£  £  £  (x  +  y)dxdydz  =  T^(c5-b5). 

I      I    (x  +  V  +  z)  d%  dy  dz  =  3  ah. 

■a%j—x  i/0 


CHAPTER   XXVI 

PLANE   AKEAS    BY   DOUBLE    INTEGKATION 
RECTANGULAR   COORDINATES 

210.  Let  y=f(x)  be  an  equation  in  which  f(x)  is  single 
valued  and  continuous  for  all  values  of  x  between  and  includ- 
ing two  values  a  and  5.  To  find  by  double  integration  the 
area  inclosed  by  the  curve  y  =/(#),  the  je-axis,  and  the  ordinates 
corresponding  to  the  abscissas  a  and  b  respectively. 


Fig.  85. 


Suppose  that  f(x)  is  positive  for  all  values  of  x  between  a 
and  b.  Under  this  supposition,  the  curve  y=f(x)  may  be  as 
in  Fig.  85. 

Let  OA  and  OB  represent  the  abscissas  a  and  b  respectively. 
Divide  AB  into  n  equal  parts.  Call  each  part  Ax.  At  Ak  and 
-4m-i>  tw0  consecutive  points  of  division,  erect  ordinates  to  meet 
the  curve  in  Bk  and,  Bk+1  respectively.  Divide  AkBk  into  m 
equal  parts.  Call  each  part  Ay.  From  each  of  the  points  of 
division  of  AkBk,  draw  a  line  parallel   to  the  #-axis  to  meet 


•Tc+l^ft+l' 


239 


240       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Denote  the  abscissa  OAk  by  xk. 

The  area  of  each  of  the  small  rectangles  is  A?/  Ax.  By  sum- 
ming the  Ay's  we  get  the  area  of  the  rectangle  AkBkCk+1Ak+1  to  be 

2j  Ay  J  Ax,  and  therefore  ^™f   2/  A2/)A^  or  (    I       efyjAx. 

Let  ajft  take  in  succession  the  values  a,  xl5  x2,  •••,  #„_!•  There- 
fore the  sum  of  the  areas  of  the  rectangles  that  may  be 
described  on  the  n  equal   parts  of   division  of   the  line  AB 


x=h 


f(xx)  f*f(x2>  /»/(*«-!) 


=    Jo     ""+/..     d2/+Jo      *+•- +J.  * 


Ax 


In  Chapter  XIX  it  was  seen  that  the  limit  of  the  sum  of 
these  areas  is  the  required  area. 


=2  1  *.)** 


Therefore  the  required  area  =  ot™^    2(    I      ^2/ )  ^ 


W  =  O0 


.  J.  •*)*■ 


»y  a  »/o 


c?x  cZ?/. 


211.   Example.     Find  by  double  integration  the  area  of  the 
circle  x2  +  y2  =  a2. 

Since  y=± Vet2  —  x2,  y  is  a  double-valued   function   of   #. 
This  difficulty  can  be   avoided  by  making  use   of   the   fact 

that  the  area  of  the  circle  is  four  times 
the  area  in  the  first  quadrant.  We 
shall  therefore  drop  the  minus  sign  and 
confine  our  attention  to  the  first  quad- 
rant. 

Make  the  same  construction  as  in  the 
Fig.  86.  general  case. 


A/A+1     x 


PLANE  AREAS  BY  DOUBLE  INTEGRATION        241 


The  area  of  the  rectangle  A,.Ck.,^  = 


limit 


/1W-+1 


y=>Ja*-x2 


'V    A?/  1  Ax 
y=o         / 

=  (J,        *)*"■ 

Let  %  take  in  succession  the  values  0,  xx,  x2,  •••,  xn^. 
Therefore  the  area  of  the  circle 


=4»=-[S(X   dyr_ 

(I  dyjdx 

n^a2— x2 
dxdy 


=  7rGr. 

EXERCISE 
If  f(x)  of  Art.  210  is  negative  for  all  values  of  x  between  a  and 

J'b     /V(x) 
I      dx  dy. 
a  Jo 

POLAR   COORDINATES 

212.  Let  r=f(0)  be  an  equation  in  which  f{6)  is  single 
valued  and  continuous  for  all  values  of  6  between  and  includ- 
ing two  values  a  and  fi.  To  find,  by  double  integration,  the 
area  inclosed  by  the  curve  ^ 

r=f(0),  and  the  radii 
vectores  that  make  angles 
of  a  and  (3  respectively 
with  the  initial  line. 

Suppose  that  the  curve 
r  =f(0)  is  as  in  Fig.  87. 

Let  AOB  and  AOC  de- 
note the  angles  a  and  /3 


respectively.     Divide  the   0 


Fig.  87 


242       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


angle  BOC  into  n  equal  parts.  Call  each  part  A0.  Denote 
the  points  at  which  two  consecutive  lines  of  division  of 
the  angle  BOC  meet  the  curve  by  Bk  and  Bk+1.  Divide  0Bk 
into  m  equal  parts.  Call  each  part  Ar.  Denote  two  consecu- 
tive points  of  division  of  OBk  by  Gt  and  Gl+i.  Denote  the  dis- 
tances of  these  points  from  0  by  rt  and  rl+1  respectively.  With 
center  0  and  radii  i\  and  rl+1  describe  arcs  of  circles  to  meet 
OBk¥1  in  H  and  L  respectively. 
Denote  the  angle  AOBk  by  9k. 

The  area  of  GlGl+iLH=th.e  area  of  OGl+1L— the  area  of  OGfl 

=  i[(n  +  Ar)2-n2]A0 

=  [r,+iAr]ArA(9. 

By  giving  r?  in  succession  the  values  0,  i\,  r2,  •  ••,  rm_l7  and 
summing  the  results,  we  get  the  sum  of  the  areas  of  all  the 
figures  formed  by  the  radii  OBk  and  OBk+1,  and  the  arcs  of 
the  circles.  That  is,  we  get  the  area  of  the  sector  OBkCk+1. 
Therefore  the  area  of  the  sector  OBkCk+l 

=  5(0  +  lAr)  +  (n  +  i  Ar)  +  (r2  +  iAr)  +  ... 

+  (rra_1  +  iAr)jArA0 

K0  +  iAr)  +  (r1-}-iAr)  +  (r2  +  iAr)  +  ... 

+  (n»-i+iAr)!ArA0~l.  (1) 


limit 
m  =  co 


Now  limit 


=  1.    Therefore,  by  the  theorem 


'(r,  + j-A^ArAf 
r;Ar  A0 

of  Art.  186,  each  term  in  (1)  may  be  replaced  by  rt  A?-  A0  where 
rt  has  the  proper  value  for  that  term.     Therefore  the  area  of 


the  sector  = 


limit 
m=cc 


jO  +  r1  +  r2+...+rTO_1jAr;A0 

/  r=X6k)  s  /    r/(ek)  \ 


PLANE  ABEAS  JiV  DOUBLE  INTEGBATION        243 


Let  6k  take  in  succession  the  values  a,  0„  6-,,  •••,  0rt  ,.    There- 
fore the  sum  of  the  areas  of  all  the  circular  sectors 


f     /VU)  /V(0,)  /'/(02) 

=  J.  r*+i   '•*•+!  '•*•+••• +1 


V(0n-1)  ) 

rdr    A6> 


In  Chapter  XX,  it  was  seen  that  the  limit  of  the  sum  of 
these  areas  is  the  required  area. 

iimit-  r^/  rf{e)      \ 

rdr)A0 


9=P  f  rm       \ 


i"  ■*  r  p  /  /*' 

Therefore  the  required  area  =  £""    V     J 

-rcr 

/•J8    /V(0) 
•/a   c/0 


reZr   tf0 


cW  f??' 


213.  Example.  Find  by  double  integration  the  area  of  the 
cardioide  r  =  2  a  (1  —  cos  0). 

Since  the  curve  is  symmetrical  to  the  initial  line,  we  may 
consider  the  part  above  the  initial  line  and  multiply  the  result 
by  2. 

Make  the  same  construction  as  in  the  general  case. 

The  area  of  the  sector 


OBkGk+1  = 


r=2a(l-cosek)  . 

f  limit        \?       rAr)A0 


V  TO  =  00 


r=0 


-a 


2a(l-cos8ic) 


rdr)\0. 


Therefore  the  area  of  the  cardioide 

r  ,.       .;   £zj/     /»2a(l-cos0) 


_  2  {  limit  y  (    f 
•/o   c/o 


rdA&o\ 


*ir    /»2a(l-cos0) 

2  I      (  rcWdr 


=  6  7r  a2 


244       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

EXERCISES 

1.  A  cornet,  C,  travels  around  the  sun,  S,  in  a  parabola,  S 
being  the  focus.  Compare  the  areas  swept  out  by  SO  as  6 
moves  from  160°  to  150°;  from  90°  to  80°;  from  10°  to  0°,  0 
being  the  angle  CSO,  where  O  is  the  vertex  of  the  parabola. 

Ans.  4541 :  29.7  :  8.77. 

2.  Solve  Exercises  23  and  24,  Chapter  XIX,  by  double 
integration. 

3.  Solve  Exercises  4  and  6,  Chapter  XX,  by  double 
integration. 

4.  In  each  of  the  remaining  exercises  of  Chapters  XIX 
and  XX,  express  the  area  in  terms  of  a  double  integral. 
Do  not  perform  the  integration. 


CHAPTER    XXVII 

VOLUMES  OP  KEVOLUTION  BY  DOUBLE  INTEGRATION 

RECTANGULAR  COORDINATES 

214.  Let  y=f(x)  be  an  equation  in  which  f(%)  is  single 
valued  and  continuous  for  all  values  of  x  between  and  in- 
cluding two  values  a  and  b.  To  find  by  double  integration 
the  volume  of  as  much  of  the  solid  generated  by  revolving 
the  curve  y  =  f(x)  about  the  a>axis  as  is  contained  between 
planes  perpendicular  to  the  &-axis  through  points  whose  ab- 
scissas are  a  and  b  respectively. 


Y 

C 

B 

k^" 

c*+TX 

D 

Gm 

i       i       i 

i 

0 

\                                        A& 

**+!                     [ 

3 

X 

Fig.  89. 

Make  the  same  construction  as  in  Fig.  85,  Art.  210  (see 
Fig.  89).  Denote  two  consecutive  points  of  division  of  AkBk 
by  G%  and  Gl+1.     Denote  the  ordinate  of  Gt  by  yv 

The  volume  generated  by  the  element  of  area  G^^F^^ 

=  7r[(yl  +  Ayy-y?]Ax 


2  7^  +  1  A?/]  A?/ Ax. 
245 


246       DIFFERENTIAL   AND  INTEGRAL   CALCULUS 


By  giving  yt  in  succession  the  values  0,  yly  y2,  y3,  •  ••,  ym_ly 

and  summing  the  results,  we  get  that  the  volume  generated  by 
the  rectangle  AkBkCk+1Ak+l 

=  2  tt[(0  +  i  Ay)  +  (?A+  J  Ay)  +  (y2  +  J  Ay)  +  ... 

+  (2/™-i  +  iAy)]AyAx 

=  m^[2-K04-|Ay)-f-(2/1  +  iAy)  +  (y2  +  jAy)  +  -.- 

+  (y«-i  +  i  Ay)}  Ay  Ax]. 


(1) 


Now 


limit 


m  =  co 


Vi+i&y 


ih 


=  1. 


Therefore,  by  the  theorem  of 

Art.  186,  each  term  in  (1)  may  be  replaced  by  y„  where  I  has 
the  proper  value  for  that  term. 

Therefore  the  volume  generated  by  the  rectangle  AkBkCk+iAk+1 

■        =^[27rJ0+y1  +  y2+...+yw_1}AyAx] 

V=/(xk) 

limit     Y    2       A 


m-- 


y=0 


Ax 


=27r(X  kydy)^x- 


Let  £c&  take  in  succession  the  values  a,  x±,  x2,  •••,  xn_v  There- 
fore the  sum  of  the  volumes  generated  by  all  the  rectangles 
described  on  the  n  equal  parts  of  division  of  the  line  AB 


=  ^(27rJ0  kyty)^x 


In  Art.  204,  it  was  shown  that  the  limit  which  this  sum 
approaches  is  the  required  volume. 
Therefore  the  required  volume 


i_. 


(2ttJo      ydyjdx 
y  dx  cly. 


VOLUMES   OF  REVOLUTION 


217 


In  the  equation  y  =  f(x),  if  a  is  a  single-valued  function  of 
y,  the  volume  of  as  much  of  the  solid  generated  by  the  revolu- 
tion of  the  curve  y  =/(#)  about  the  ?/-axis,  as  is  contained 
between  planes  perpendicular  to  the  ?/-axis  through  the  points 
on  the  curve  whose  ordinates  are  c  and  d  respectively  can  be 
found  in  a  similar  manner  to  be 


=  2.f  f 


x  dy  dx. 


215.  Example.  Find,  by  double  integration,  the  volume 
generated  by  the  revolution  about  the  cc-axis  of  as  much  of 
the  parabola  y2  =  2  x  as  is  contained  between  the  origin  and 
the  plane  perpendicular  to  the  a>axis  through  a  point  whose 
abscissa  is  4. 

The  curve  is  as  in  Fig.  90. 

Make  the  same  construction  as  in  Fig.  85,  Art.  210. 

The  volume  of  the  cylinder  generated  by  the  revolution  of 
the  rectangle  AkCk+1 


=^sr 


V  dy)kx. 


Therefore  the  required  volume 
=  2  7r  I     I      xy  dxdy 


x  dx 


Fig.  90. 


=  16 


7T. 


POLAR  COORDINATES 

216.  Let  r=f(0)  be  an  equation  in  which  f(0)  is  single 
valued  and  continuous  for  all  values  of  0  between  and  includ- 
ing two  values  a  and  ft.     To  find  the  volume  generated  by  the 


< 


248       DIFFEBENTIAL  AND  INTEGRAL   CALCULUS 

revolution  about  the  initial  line  of  the  radii  vectores  that  make 
angles  of  a  and  ft  respectively  with  the  initial  line,  and  the 
part  of  the  curve  included  between  these  radii  vectores. 

Suppose  that  f{6)  is  positive  for  all  values  of  6  between  a 
and  (3. 

Under  this  supposition,  the  curve  may  be  as  in  Fig.  91. 

Make  the  same  construction  as  in  Fig.  87,  Art.  212. 


Fig.  91. 


The  volume  generated  by  revolving  the  area  GtGl+iLH  about 
the  initial  line  is  greater  than  2  7rrt  sin  0A[r,  Ar  A0  -f  \  Ar2  A0], 
and  less  than  2  7r(r,+  A?')  sin  (6k  +  A0)[r,  Ar  A<9  +  |  Ar2  A0]. 
Call  it  <f>(rv  6k)Ar  A0. 

By  giving  rt  in  succession  the  values  0,  rl5  r2,  •••,  rm_13  and 
summing  the  results,  we  get  that  the  volume  generated  by  the 
sector  OBkOk+1 

=       5)  <f>(r,  0fc)Ar  |  A0 

*-       r=0 


limit 
m  =  co 

r=0 


]T  <£(r,  0,)Ar    A0 


£Bk)<t>(r,Ok)dr 


VOLUMES   OF  REVOLUTION 


249 


Let  0k  take  in  succession  the  values  a,  0V  02,  03>  ■■•>  0«-i- 
Therefore  the  sum  of  the  volumes  generated  by  the  circular 
sectors  that  may  be  described  on  the  n  equal  parts  of  division 


of  the  angle  BOC 


2(j[      *(r,0)drJAft 


Since  the  sum  of  these  circular  sectors  approaches  the  given 
area  as  its  limit  as  n  increases  without  limit,  the  sum  of  the 
volumes  generated  by  these  sectors  ajjproaches  the  required 
volume  as  its  limit  as  n  increases  without  limit. 

Therefore  the  required  volume 


=^\%r^^¥6 


=  f  Y  Cfi%(r,  0)  dr\  d6 


=  (  cj>(r,  6)  dO  dr. 

J  a    Jo 


Now  2  7T  r{  sin  O^n  Ar  A0  +  l  Ar2  A0] 

<4>(r9  0&)Ar  A0<2  B-(r,  +  A?-)  sin  (04  +  A0)  [r,  A?-  A0  +  J  A?'2  A0]. 
Divide  by  2  ?r  r?2  sin  0^.  A?-  A0. 


r,  Ar  A0 


< 


2  7r  ?'j2  sin  0* 


(r,  +  Ar)  sin  (0k  +  Afl)  [r,  Ar  A0  +  \  Ar2  A0] 
r£2  sin  0&  Ar  A0 

Since  the  limit  of  each  extreme  of  this  inequality  as  m  and 
n  both  become  infinite  is  1, 


limit 
m  =  co 
%  =  cc 


'    <ftfo,0*) 

2  7r  i\  sin  < 


=  1. 


250       DIFFERENTIAL  AND  INTEGRAL    CALCULUS 

Therefore,   by    the    theorem    of    Art.    186,   each    term    in 

X/3     /V(0) 
<l>(r,  6)d0  dr  may  be  replaced  by  2  tt  r?  sin  6k  A0  A?-,  where 

k  and  I  have  the  proper  values  for  that  term. 

Therefore  the  required  volume  =  2  tt  I      I       r2  sin  0  d0  dr. 


EXERCISE 


In  r  =/(0)  of  Art.  216,  if  f(6)  is  negative  for  all  values  of  0 
between  a  and  /?,  show  that  the  area  denned  in  that  article  is 

•jB   /V(0) 


I  ""  r2  sin  0c20  dr. 


217.  As  an  illustration  of  the  application  of  the  theorem 
of  the  preceding  article  to  problems  in  volume,  consider  the 
following  example. 

Example.  Find  the  volume  generated  by  the  revolution 
about  the  initial  line  of  the  cardioide  r  =  2  a(l  —  cos  0). 

The  volume  of  the  sector  OBkCk+1 


Fig.  92. 


2a(l—  cos0 


%(r,0*)drW 


Therefore  the  required  volume 

S*n    /»2o(l— cos 


Jr"7r     /»J 
I 
0    Jo 


7^  sin  QdOdr 


64  _^3 
3 


=  V  Ira?. 


EXERCISES 


1.   Find,  by  double  integration,  the  volume  generated  by 

2  2 

the  revolution  of  the  ellipse  —  +  —*  =  1  about  the  a>axis  ;  about 


the  ?/-axis. 


62 


^4ws.  f  7ra&2;  f  7ra26. 


2.   Find,  by  double  integration,  the  volume  generated  by  the 

2  2  2-1 

revolution  of  the  four  cusped  hypocycloid  xz  +  ys  —  as  about 
the  a>axis.  Ans.  f\25  v  a*- 


VOLUMES   OF  REVOLUTION  251 

3.  Find,  by  double  integration,  the  volume  generated  by 
the  revolution  of  one  arch  of  the  cycloid  x  =  a{6  —  sin  0), 
y  =  a(l  —  cos  0)  about  the  #-axis ;  about  the  ?/-axis. 

Ans.  5  7r2  a8 ;  6  it'  a3. 

4.  Find  the  volume  generated  by  the  revolution  of  the  circle 
r  =  a  about  the  initial  line.  Ans.  \  tt  a\ 

5.  Find  the  volume  generated  by  the  revolution  of  one  loop 
of  the  lemniscate  r  =  a  cos  2  0  about  the  initial  line. 

Ans.  ?^[8V2-91. 
105  L  J 

6.  Find  the  volume  generated  by  the  revolution  of  one  loop 
of  the  lemniscate  r  =  a  sin  2  6  about  the  initial  line. 

a        32       3 

Ans.  7r  aJ. 

105 

7.  Find  the  volume  generated  by  the  revolution  of,  (a)  the 

small  loop  of  the  curve,  (6)  the  whole  curve  r  =  a  sin  i  0  about 

the  initial  line.  2V2      ,     2 

Ans.    7ra6:  ■ — 

15  '  15 


23-V2 


7ra° 


8.  Find  the  volume  generated  by  the  revolution  of,  (a)  the 
loop  in  the  first  quadrant,  (b)  the  loop  below  the  initial  line 
of  the  curve  r  =  a  sin  3  6  about  the  initial  line. 

A       27 'V3       3    27 V3      3 


CHAPTER   XXVIII 


VOLUMES.     SUKFACES 


218.  Let  z  =f(x,  y)  be  the  equation  of  a  surface  that  cuts 
the  #?/-plane,  the  #2-plane,  and  the  planes  whose  equations  are 
x  =  a  and  x  =  b  in  curves  which  together  inclose  a  region  on 
the  surface. 

Suppose  that  fix,  y)  is  single  valued  and  continuous  for  all 
values  of  x  and  y  which  together  determine  a  point  in  the 
region  of  the  cw/-plane  bounded  by  the  curve  of  intersection  of 
the  surface  and  a??/-plane,  and  the  lines  whose  equations  are 


Fig.  93. 


x=  a  and  x—b.  To  find  the  volume  inclosed  by  the  ajy-plane, 
the  072-plane,  the  planes  whose  equations  are  x  —  a  and  x  =  b, 
and  the  given  surface. 

Suppose  that  the  surface  is  as  in  Fig.  93. 

252 


VOLUMES  253 

Let  OA  and  OB  represent  a  and  b  respectively.  Divide  AB 
into  n  equal  parts.  Call  each  part  Aa?.  At  Ak  and  ^4ft+1  draw- 
planes  perpendicular  to  the  a>axis  to  meet  the  surface  in  LKFBh 
and  MHGBk+l  respectively.  Divide  AkBk  into  m  equal  parts. 
Call  each  part  Ay.  At  Ct  and  Cl+l  draw  planes  perpendicular 
to  AkBk  to  meet  the  surface  in  KHaxid  FG  respectively.  Divide 
CtK  into  p  equal  parts.  Call  each  part  Az.  At  each  of  the 
points  of  division  of  CtK  draw  a  plane  perpendicular  to  GZK. 

Denote  the  abscissa  0Ak  by  xk.  Denote  the  ordinate 
4bQ  by  y, 

In  z  =f(x,  y),  let  2;  =  0  and  we  have  the  equation  of  the  curve 
CBkBk+lD.  Solve  f(x,  y)  =  0  for  y.  Suppose  that  the  result  is 
y  =  <f>  (x).     Then  the  coordinates  of  Bk  are  (xk,  <f>(xk),  0). 

The  volume  of  each  of  the  parallelopipeds  formed  by  the  six 
planes  is  Az  Ay  Ax.  By  summing  the  Az's  from  Ct  to  Kwe  get 
that  the  volume  of  the  rectangular  parallelopiped  Fl+1K 

z=f(xk,yi)  z=f(xk,yi) 

or  (I  d^jAy  Aa;. 

This  is  the  volume  of  a  parallelopiped  one  of  whose  corners, 
C\,  is  distant  yj  from  the  a>axis.  Let  yl  take  in  succession  the 
values  0,  yp  y2,  y3,  •••,  ym_v  Then  the  sum  of  the  volumes  of 
all  the  rectangular  parallelopipeds  that  can  be  described  on  the 
m  equal  parts  of  division  of  the  line  AkBk  is 


Ay  Ax, 


\X    dz+X    dz+X     *+•••+ i      * 

=  1 S  C Jo   *)**}-**  - 

The  limit  which  this  sum  approaches  as  m  increases  without 
limit  is  the  volume  of  the  solid  bounded  by  the  x?/-plane,  the 
ajz-plane,  the  planes  whose  equations  are  x  —  xh  and  x  =  di 


k+ly 


25 J:       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

and  the  cylindrical  surface  one  of  whose  bases  is  the  curve 
LKFBk. 

Then  the  volume  of  the  solid  just  described 


=  {Jo        (Jo  *)«*}*»■ 

This  is  the  volume  of  the  solid,  one  of  whose  corners,  ^.,  is  at 
a  distance  scj.  from  0.     Let  ^  take  in  succession  the  values  a,  xx 
x2,  '"}  xn-v    Then  the  sum  of  all  the  solids  that  can  be  described 
on  the  n  equal  parts  of  division  of  the  line  AB  is 

li   U    *>+J[    (J.    <**>+./.    (1    *> 


It  can  be  shown  by  a  more  or  less  elaborate  investigation 
which  will  not  be  entered  into  here,  that  the  limit  of  this  sum 
as  n  increases  without  limit  is  the  required  volume.  Assuming 
that  this  limit  is  the  required  volume,  we  have : 

The  required  volume  =  ^™  V  |    I        (I         dzjdy  \  Ax 

x=a    "■  *  '  ■* 

l  tJ.  y.  *>•}*• 

=  ||  I         dxdydz. 

EXERCISE 

Determine  the  expressions  for  the  volume  of  Art.  218  when 
it  lies  in  the  other  octants. 


VOLUMES 


255 


219.    As  an  illustration  of  the  application  of  the  theorem  of 
the  preceding  article  to  prob- 
lems in  volume,  consider  the 
following  example. 

Example.      Find   the   vol- 
ume of  the  ellipsoid 

0  0  0 

a2      b2      c2 

Consider   the   part    of    the 
ellipsoid  in  the  first  octant. 

The  surface  of  this  much  of    y 
the  ellipsoid  is  as  in  Fig.  94. 

Make  the  same  construction  as  in  the  general  case. 

The  volume  of  the  parallelopiped  Fl+1K 


Fig.  94 


2  =  0 


,cV 'i-^2  -VJ-. 


Therefore  the  volume  of  the  solid  bounded  by  the  xy  plane,  the 
coz-plane,  the  planes  whose  equations  are  x=xk  and  x=xk+1,  and 
the  cylindrical  surface  one  of  whose  bases  is  the  curve  LKFBk 


y-h 


I    limit  "sr^ 
1  m  =  co  Z-i 

V  y=0 


X    •  l'd*YA 


Ax 


=  iX    ""(X       *)* 


Ax. 


Therefore  the  required  volume 

=tesl   IX      d¥ 


x=0 


] 


=  8  I      I  I  dxdydz 

c/0     Jo  Jo 


r-JTjf  W-Sh* 


256       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


VHP 
1 cos  d>. 
ar 


'       x2 


:  dy  =  —  b\jl -sin<£d</>.     Also,  <j>  =  ^  when  y=0,  and 

*  CI  u 


ch  =  0  when  y  =  b\ll  —  — • 


=  —  8  6c  I      I    f  1  —  —  ]  sin2  cf>  clx  dd>. 
Jo  J%  \        a2  J 

Therefore  the  required  volume  =  2  tt  be  I    j 


1 5  )  tfec 

a2 


=  1 7r  abc. 


SURFACES 

220.  Let  2  =f(x,  y)  he  the  equation  of  a  surface  that  cuts 
the  icy-plane,  the  #2-plane,  and  the  planes  whose  equations  are 
x  =  a  and  a;  =  b  in  curves  which  together  inclose  a  region  on 
the  surface.  Suppose  that  f(x,  y)  is  single  valued  and  con- 
tinuous for  all  values  of  x  and  y  which  together  determine  a 
point  in  the  region  of  the  x?/-plane  bounded  by  the  curve  of 
intersection  of  the  surface  and  a??/-plane,  and  the  lines  whose 
equations  are  x  =  a  and  x=b.  To  find  the  area  of  the  surface 
bounded  by  the  scy-plane,  the  jcz-plane,  and  the  planes  whose 
equations  are  x  =  a  and  x  =  b. 

221.  Before  proceeding  to  a  consideration  of  the  problem 
stated  in  the  preceding  article,  we  shall  first  show  that  the 
angle  y  which  the  tangent  plane  to  the  surface  z=f(x,  y) 
makes  with  the  ojy-plane  at  a   point  whose  coordinates  are 


fe,  H*  is  such  that   Secy  =  (Vl  + 


§A2,fdz\2 

dxj      \dy 


y=y0 


SURFACES 


257 


Fig.  95. 


Let  P  with  coordinates  (x0,  y0,  z0)  be  a  point  on  the  surface. 
(Fig.  95.)  Through  P  draw  a  tangent  plane  to  the  surface  and 
also  planes  parallel  to  the  coor- 
dinate planes.  Let  PM  =  Asc, 
and  PN=Ay.  Through  M 
draw  a  plane  parallel  to  the 
yz-plane,  and  through  N  draw 
a  plane  parallel  to  the  #2-plane. 
Let  these  planes  meet  the  tan- 
gent plane  at  P  in  lines  on 
which  are  the  points  M'  and 
N1    respectively,    as    indicated. 

Through  P  draw  the  line  TPS  perpendicular  to  the  tangent 
plane. 

Let  MPM'  =  <f>,  and  N'PN=\p.  Then  the  direction  cosines 
of  PM'  are  cos  <£,  0,  and  sin  <£,  and  of  PN'  are  0,  cos  if/,  and 
sin  if/.  Let  the  line  TPS  have  the  direction  angles  a,  fi,  and  y. 
The  angle  y,  since  it  is  the  angle  between  the  lines  perpen- 
dicular to  the  tangent  plane  and  the  a??/-plane  respectively,  is 
the  angle  between  the  two  planes. 

Since  TPS  is  normal  to  the  tangent  plane  it  is  perpendicular 
to  PM'  and  PN'. 

.'.  cos  a  cos  <ft  +  cos  y  sin  <f>  =  0, 


and 


cos  /?  cos  if/  -f-  cos  y  sin  if/  =  0. 


Substitute  in  cos2  a  +  cos2  /?  -f  cos2  y  =  1. 

. \  cos2  y  tan2  <£  +  cos2  y  tan2  ^  +  cos2  y  =  1. 

.*.    COS    v  = 

'      l  +  tan2<£  +  tan2^ 
.\  sec2  y  =  1  +  tan2  <f>  -f-  tan2  if/. 


.:  sec  y  =  Vl  +  tan2  <j>  -f  tan2  ^. 


258       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


By  Art.  127,  tan  <£  = 


dx 


,  and  tan  \p  = 


by 


secy=M 


1+(S)" 


+ 


dz\ 


x — a;0 
i/=y0 


x—x0 

y=y0 


222.    To  return  to  the  problem  of  Art.  220. 

Suppose  that  the  surface  is  as  in  Fig.  96.  Let  OA  and  OB 
represent  a  and  b  respectively.  Divide  AB  into  n  equal  parts. 
Call  each  part  Aaj.  At  Ak  and  Ak+l  draw  planes  perpendicular 
to  the  i^axis  to  meet  the  surface  in  BkL  and  Bk+1M  respectively. 


Fig.  96. 

Divide  AkBk  into  m  equal  parts.  Call  each  part  Ay.  At  Ct  and 
Gl+X  draw  planes  perpendicular  to  the  line  AkBk  to  meet  the 
surface  in  KN  and  FG  respectively.  At  K  draw  the  tangent 
plane  to  the  surface.  Denote  the  angle  which  this  tangent 
plane  makes  with  the  #?/-plane  by  y. 

Denote  the  abscissa  OAk  by  xk.     Denote  the  ordinate  AkCt 
by  yv 


SURFACES 


259 


In  z  =  f(x,  y),  let  z  =  0  and  we  have  the  equation  of  the 
curve  CBl;Bk^D.  Solve  f(x,  y)  =  0  for  y.  Suppose  that  the 
result  is  y  =  <f>  (x).    Then  the  coordinates  of  Bk  are  (xk,  <f>(xk),  0). 

The  area  of  the  tangent  plane  cut  off  by  the  four  planes  C\F, 
CXN,  Cl+1G,  and  FtG  is 


sec  y  Ay  Ax,  or 


(V 


•■(• 


dz\2  ,  /dz\ 


+fe; 


y=y0 


Ay  Ax. 


Let  yt  take  in  succession  the  values  0,  ?/1?  ?/2,  •-.,  ym_l.  Then  the 
sum  of  the  areas  of  all  the  tangent  planes  that  can  be  de- 
scribed like  the  above  one  on  the  m  equal  parts  of  division 
of  the  line  AkBk 


[(^K!^!■L,.+(^' 


™H^(iJ+(U)-,+ 

y=y0  y=v. 


-(VKlHi) 


—  it 


Ay  Ax, 


or 


L    y=0        \ 


dz 
dx 


+ 


dyj 


Ay 


Ax. 


The  limit  which  this  sum  approaches  as  m  increases  without 
limit  is  the  area  of  as  much  of  the  curved  surface  formed  by  a 
tangent  line  to  the  given  surface  which  moves  with  its  point 
of  contact  on  the  curve  LKFBk  as  is  contained  between  the 
planes  whose  equations  are  x  =  xk  and  x  =  xk+1.     Then  the  area 

*V. /-,  .  Tew  ,  /m2\        -i 

Ax, 


of  the  surface  =  I   l™*    ^    fyjl+f™\ 


\2      fdz^2 


Ay 

x=xk  _ 


HW.<% 


dy  j  Ax.     Let  xk  take  in  suc- 


cession  the  values  a,  xlt  x2,  •-,  xn_x.  Then  the  sum  of  the 
areas  of  all  the  surfaces  described  above  that  can  be  described 
on  the  n  equal  parts  of  division  of  the  line  AB 


260       DIFFERENTIAL  AND  INTEGRAL    CALCULUS 


■vr(W$<? 


dy 


+ 


riMtHH)' 


dy 


+  •••  + 


srxHS)'+(MJ 


dy 

x==xn — 1       


Ax 


■Krv^+^v 


dyJ    J 


It  can  be  shown  by  an  investigation  which  would  be  beyond 
the  scope  of  this  book  that  the  limit  which  this  sum  approaches 
as  n  increases  without  limit  is  the  area  of  the  given  -surface. 
Assuming  that  this  limit  is  the  area  of  the  given  surface,  we 
have  that  the  area  of  the  given  surface 


-;:"[S(fV 

I    x=a    » 


I+'£M«h 


Ax 


223. 


=£TV>+(IH§)W 

Example.     Find  the  area  of  the  surface  of  the  sphere 

x2  +  y2  +  z2  =  a2- 

Consider  the  part  of  the 
sphere  in  the  first  octant. 

Make  the  same  construction 
as  in  the  general  case. 


Since        z  =  Va2  —  x2  —  y2, 


—  x 


dx     Va2-a2-2/2 


-y 


Fig.  97 


dy      -yjtf-tf  —  y2 


SURFACES 


261 


••  MiKiM* 


ur 


z  +  ^ 


a"  --  x2  —  y2  '  a2  —  x2  —  y2 


a 


Va2  —  x2  —  if 


The  area  of  as  much  of  the  curved  surface  formed  by  a 
tangent  line  to  the  given  surface  which  moves  with  its  point 
of  contact  on  the  curve  LKBk  as  is  contained  between  the 
planes  whose  equations  are  x  =  xk  and  x  =  xk+1 


_  f  limit  y=v^-'*2         a  Ay 


m  =  <x> 


X 

y=0 


Va2- 


Aoj 


=7  f^7_    «<%      VAa, 

Vy°  ^/a2-x2-y2J 

Therefore  the  required  area 

__  8  ["limit  ^y  /  r^^72 


a  dy 


Va2  —  x2  —  y2, 
Jo  Jo 


Ax 


Va2  —  x2  —  y2 


=  4  7T  a2. 


EXERCISES 


1.    Find  the  volume  inclosed   by  the  icy-plane,   the  plane 


z  =  mx,  and  the  cylinder  x2-\-y2  =  a2. 


Ans.  -|  a3m. 


2.   Find  the  volume  common  to  the  two  cylinders  x2  +  y2  =  a< 


and  x2  +  z2  =  a2. 


^4?2s.  J^-  a3. 


3.    Find  the  volume  common  to  the  paraboloid  of  revolution 
y2  +  z2  =  4  a#  and  the  cylinder  a,*2  +  ?/2  =  2  ace. 


-<4?is. 


2  7T  4" 


16" 


a,J 


262       DIFFERENTIAL  AND  INTEGRAL    CALCULUS 

4.  Find  the  volume  inclosed  by  the  surface  (-]   -{-(-) 

-f  ( - )  =1  and  the  positive  sides  of  the  three  coordinate  planes. 

W  .        abc 

Ans.  — —  • 
90 

5.  Find  the  area  of  the  surface  cut  off  from  the  sphere 
x2  +  y2  +  z2  =  a2  by  the  cylinder  x2  +  y2  =  ax.      Ans.  2  [7r  —  2]  a2. 

6.  Find  the  area  of  the  surface  cut  off  from  the  cylinder 
x2  -\-  y2  —  ax  by  the  sphere  x2  +  y2  +  z2  =  a2.  Ans.  4  a2. 

7.  Find  the  area  of  the  surface  inclosed  by  the  two  cylinders 
x2-\-y2  =  a1  and  x2  +  z2  =  a2.  Ans.  16  a2. 

8.  Find  the  area  of  the  convex  surface  described  in  Ex- 
ercise 1.  Ans.  4  a2m. 

2.  2  2 

9.  The  equation  of  the  base  of  a  cylinder  is  ys  -f  z3  =a3 
and  its  axis  is  parallel  to  the  #-axis.  Find  the  area  of  the  por- 
tion of  the  surface  bounded  by  a  curve  whose  projection  on  the 
ajy-plane  is  x3  +y^  =  a3.  Ans.  -U  a2. 

10.  A  square  hole  is  cut  from  a  sphere,  the  axis  of  the  hole 
coinciding  with  a  diameter  of  the  sphere.  The  radius  of  the 
sphere  is  a  and  a  diagonal  of  the  hole  is  2  V2  b.  Find  the  area 
of  the  surface  cut  from  the  sphere  by  the  hole. 

r-l  TO 

2  b  sin-1  —        '     —  a  sin-1 


.[S 


Va2  — 62  ar-b2_ 


CHAPTER   XXIX 

SOME  METHODS   OP   APPROXIMATE  INTEGRATION 

224.  In  the  problems  thus  far  considered,  involving  the 
definite  integral  of  an  expression,  the  definite  integral  could 
always  be  expressed  as  the  difference  between  two  values  of  an 
indefinite  integral.  In  many  important  cases,  however,  the 
definite  integral  cannot  be  so  expressed.  When  such  a  case 
arises,  we  are  usually  content  with  an  approximate  value  of  the 
definite  integral,  found  by  developing  the  integrand  into  a 
power  series  by  Taylor's  Theorem  or  one  of  its  special  cases, 
Maclauren's  or  the  Binomial  Theorem,  and  integrating  the  first 
few  terms  of  the  series  between  the  given  limits.  Two  cases 
where  approximate  values  can  be  found  in  this  manner  will  now 
be  investigated.  From  their  importance  they  are  given  special 
names, — the  Elliptic  Integrals  of  the  First  and  Second  Classes 
respectively. 

Case  I.  It  will  be  shown  later  that  in  a  pendulum  of  length 
a,  if  the  bob  starts  from  the  lowest  point  of  its  path  with  the 
velocity  which  it  would  acquire  if  it  fell  freely  in  a  vacuum 
through  a  distance  h,  the  time  it  takes  to  reach  a  point  whose 
ordinate  is  y  is  a      ~y  d 

f27jJ°  -y/{h-y)(2ay-fj 


t  = 


r      a  (i) 


where  the  equation  of  the  path  is  x2  +  y2  —  2  ay  =  0. 
If  h  <  2  a,  let  x2  =  ^-,  and  (1)  becomes 


fa  C° 


dx 


V^X'-sV) 


263 


264       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

where  the  integral  is  of  the  form 

dx 


f: 


V(i-^)(i-fe2) 

k2  being  positive  and  less  than  1. 

If  7i  >  2  a,  let  x2  =  -f- ,  and  (1)  becomes 
2  a 


a*qlJo 


dx 


where  the  integral  is  of  the  form 

Cx dx 

J°  ^(l-x2)(l-k2x2) 

k2  being  positive  and  less  than  1. 

Case  II.     It   can   be   readily   shown   by    the    methods   of 
Chapter  XXII,   that   the   length   of   the   arc   of   the   ellipse 

— \-  *-  =  1  contained  between  the  point  (o,  b)  and  the  point 
a2     b2 

(x,  y)  on  the  curve  is 


where  e  is  the  eccentricity  of  the  ellipse. 
Replace  -  by  x,  and  (1)  becomes 


~X  Vfe?  dx>  (1) 


o2^2 


s=a£^ 


X2 

where  the  integral  is  of  the  form 


Jo    \  i_-« 


X 

k2  being  positive  and  less  than  1. 


METHODS  OF  APPROXIMATE  INTEGRATION       265 

Jr%x  dx 

,  is 

o   V(l-a;2)(l-A;2^) 

called  an  Elliptic  Integral  of  the  First  Class.     It  is  denoted  by 

F(k,  x).  

s*x    R  TePx2 

The  expression   |    \\- -dx  is  called  an  Elliptic  Integral 

c/o     *  1  —  ar 

of  the  Second  Class.     It  is  denoted  by  E(k,  x). 

The  expressions 


are  called  the  Complete  Elliptic  Integrals  of  the  First  and 
Second  Classes  respectively.  They  are  usually  denoted  by 
K  and  E  respectively. 

226.  The  substitution  of  x  =  sin  <j>  in  the  Elliptic  Integrals 
reduces  them  to  the  forms 

F(k,  <£)=  f*  d*    ,aiidm^)=  fVl-^sin^cfy 

*^°  Vl  —  k2  sin2  *  J° 

respectively. 

227.  An  approximate  value  of  F(7c,  *). 

Develop  (1  —  A;2  sin2  <j>)~z  into  an  infinite  series  by  the 
Binomial  Theorem. 

Since  k  is  less  than  1,  k2  sin2  *  is  less  than  1  for  all  values 

Of*-  1      o 

.-.  (1  -  k2  sin2  <£)-*  =  1  +  1  k2  sin2  <£  +  ±_|  A;4  sin4  «£  +  ...  (1) 
for  all  values  of  <f>.     (See  Art.  95.)  '— 

If  the  right-hand  member  of  (1)  be  integrated  term  by  term 
with  respect  to  cf>  between  the  limits  0  and  <£,  the  resulting 
series  is 

7,2 

<ft-\ —  (<f>  —  sin  <£  cos  <f>)  —  -gs  ^4  s^n3  *  cos  * 

+  &#(*- sin*  cos  *)....       (2) 


266       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Series  (2)  gives  an  approximate  value  of  F(k,  <f>)  for  given 
values  of  k  and  cf>,  the  degree  of  accuracy  depending  on  the 
number  of  terms  computed  in  the  series. 

228.    Example.     Find  an  approximate  value  of  F' 


,  2  '6 

For  k——^-,  <£=^j  series  (2)  becomes 

""  ■  1f7r     ^V    3  .1.1.  V3  ,  I  .  lAr     V3\ 


6  +  S\6~TJ~ 32  '  4  '  8  '  ~2~  +  64    4V6       4  y 

=  0.52359 ...  +  0.01132 0.00253  •••  +  0.00318  ••• 

=  0.53556.-.. 

An  approximate  value  of  F[  —— ,  5  )  is  therefore  0.53556. 

\2      6J 

Note.  The  approximate  values  of  this  chapter  are  not  approximate  in 
the  sense  that  they  are  correct  to  the  number  of  decimal  places  given. 
They  are  merely  results  found  in  the  attempt  to  get  the  actual  values  of 
the  integrals. 

229.   An  approximate  value  of  E(k,  cf>). 

Develop  (1  —  ft2  sin2  <£)*  into  an  infinite  series  by  the  Binomial 
Theorem. 

.-.  (l-Fsin2«/>)^  =  l-lA;2sin2^-li^A;4sin^  +  ...         (1) 

for  all  values  of  <£. 

Integrate  each  term  of  the  right-hand  member  of  (1)  with 
respect  to  <£  between  the  limits  0  and  <£.  The  resulting 
series  is 

</> (<£  —  sin  <£  cos  (/>)  +  —sin3  <£  cos  <£ 

_  ^k4(cf>  -  sin  <£>  cos  <j>)  +  ••••       (2) 

Series  (2)  gives  an  approximate  value  of  E(k,  <f>)  for  given 
values  of  k  and  <£,  the  degree  of  accuracy  depending,  as  in  the 
case  of  F(k}  cf>),  on  the  number  of  terms  computed  in  the  series. 


METHODS   OF  APPROXIMATE  INTEGRATION      267 

/V2      7TN 


230.    Example.     Find  an  approximate  value  of  E\  — ,  - 

For  k  —  -¥— ,  4>  =7^,  series  (2)  becomes 
2  6 


7T 


1/V      V3\  ,1      1     1     V3      3      1/V      V3 


V1 

y    32 


6      8V6        4/32     4     8       2        64     4^6        4 

=  0.52359 0.01132...  +  0.00084 0.00106-  ■ 

=  0.51205..-. 
An  approximate  value  of  El  — -,  -  J  is  therefore  0.51205. 

231.  Find  the  length  of  the  arc  of  the  lemniscate  r*—o?  cos  2  6 
contained  between  the  point  (a,  0),  and  the  point  (r,  0)  on  the 
curve. 


for  ? 


Since  ?-2  =  a2  cos  2  0,  .-.  r  =  ±  a Vcos  2  0.     Take  the  -f  sign 
c?r      a  sin  2  0 


tf0      Vcos  2  6> 


Jr2  +  ^Y=Ja2cos20  +  a2sin22^=— ^ 

V         W      A  T    cos20        Vc^s~20 

Jo   v  cos  2  0 


=  aJo  vr^2~ 


2sin20 

This  is  not  an  Elliptic  Integral,  since  Jc2,  or  2,  is  not  less  than 
1.  It  can,  however,  be  put  in  such  a  form  by  means  of  the 
transformation  2  sin2  6  =  sin2  cj>. 

a/2 
Let  2  sin2  0  =  sin2  d>.    .-.  sin  0  =  -^—  sin  <f>. 


V2  r^       ^<fr 

2  Jo   Vl-!-sin2<£ 


268       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


For  given  values  of  <f>,  s  can  be  found  approximately  in  terms 
of  a  by  the  method  of  Art.  227. 

232.    Find  the  length  of  the  arc  of  the  hyperbola  —  —  ^-  =  1 

a2      b2 

contained  between  the  point  (a,  0),  and  the  point  (x,  y)  on  the 

curve. 

Since  --^-=1,  .-.  x  =  ±  J  vV  +  b2.     Take  the  +  sign  for 
cr      b-'  b 

dx     a 


x. 


y 


dy     b  V2/2  +  b2 


•wi+(gHa 


9  9 


1  + 


b2{y2  +  b2) 

(a2  +  b2)y2^ 
b4  ' 


V(a2+b2)y2+b4 
Vb2y2  +  64 


1  + 


£ 


■■-jr 


f  1     -aVy8  1 


62 


cty, 


where  e  is  the  eccentricity  of  the  hyperbola. 


aey  _ 


In  (1),  let  ^f  =  tan  <f>. 


b2  r 
.-.*  =  —  I 

aeJo 


b2  rt>     sec2cj>dcj> 


^5 


& 


Since  sec2  <f>  =  1  +  tan2  <£, 


62  f « 

*  aeJo 


c?<£ 


b2  C*    tan2<£c7<£ 


^/l         Sin24> 


x 


ae      \e     J      aeJo 

V1 


62   /•*    tan2<M4> 


2 


sur  <£ 

e2 


a) 


(2) 


METHODS   OF  APPROXIMATE  INTEGRATION       2G9 

Now  h-  f    tan^(^ 

^l--sm2<£ 

f — surd)  H -snr  <f>-l-tarr  d>  «<£ 

-  &!  M     62  ^/>2     y     //     b2      y  y 

«eJo  r      1    . 

^--sin2^ 

.[  1 -sin2<&yM 

Jo  I 1 

V1_^sin2<jf> 

fl-Isin2^  +  -^-9tan2^ 

X<t\       e~  we-  ) 



%M 


sm^ 


fl--sin2d>  +  - 
M        e2        ^ 

'         i 

Jo  1 


=  —  ae  E  [  -,  cf>  ]  +  ae 


V-isin^ 


=  —  ae  J5J  f  -j  <£  J  -f-  ae  tan  <£\/l  — -  sin2  cf>, 


e' 
since  the  last   integral   becomes  j  *W  on  the  substitution  of 


u  =  tan  </>\/l -2  sin2<£. 


.*.  s  =  —  F[  -,  (f>)  —  aeEf-,^}  -f-aetan  <£\/l -9  sin2<£. 

ae     \e     J  \e     J  *        el 

233.  Tables  giving  the  values  of  the  Elliptic  Integrals  of  the 
First  and  Second  Classes  to  ten  places  of  decimals  are  given 
in  Legendre's  Traite  cles  Fonctions  Elliptiques.  In  the  next 
article  is  given  a  small  three-place  table  compiled  from  these 
which  may  be  used  in  solving  the  following  exercises. 


( 


270       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


EXERCISES 

1.  Compare  the  results  found  for  F{^A j  A  and  e(^A \  1 
in  Arts.  228  and  230  with  those  found  by  using  the  tables. 

cc  1J 

2.  Find  the  lengths  of  the  arcs  of  the  ellipses \-  —  =  1 

16     25 

9  9 

and h  f-  =  1. 

36     16 


Ans.  28.32:  31.68. 


3.  Find  the  length  of  the  arc  of  the  lemniscate  given  in 
Art.  231.  Ans.  5.243  a. 

4.  Find  the  length  of  the  arc  of  the  hyperbola —  =  1 

contained  between  the  points  (4,  0)  and  (8,  3V3). 

Ans.  6.725. 


Art.  234. 


F(k,  <£) 


£=0 

£  =  0.1 

£  =  0.2 

£  =  0.3 

£=0.4 

£  =  0.5 

£  =  0.6 

£  =  0.7 

£  =  0.8 

£  =  0.9 

£  =  1 

4> 

sin0° 

sin  6° 

sin  12° 

sinlS° 

sin  24° 

sin  30° 

sin  37° 

sin  45° 

sin  53° 

sin  64° 

sin  90° 

0° 

0.000 

0.000 

0.000 

0.000 

0.000 

0.000 

0.000 

0.000 

0.000 

0.000 

0.000 

5° 

0.087 

0.087 

0.087 

0.087 

0.087 

0.087 

0.087 

0.087 

0.087 

0.087 

0.087 

10° 

0.175 

0.175 

0.175 

0.175 

0.175 

0.175 

0.175 

0.175 

0.175 

0.175 

0.175 

15° 

0.262 

0.262 

0.262 

0.262 

0.262 

0.263 

0.263 

0.263 

0.264 

0.264 

0.265 

20° 

0.349 

0.349 

0.349 

0.350 

0.350 

0.351 

0.352 

0.353 

0.354 

0.355 

0.356 

25° 

0.436 

0.436 

0.437 

0.438 

0.439 

0.440 

0.441 

0.443 

0.445 

0.448 

0.451 

30° 

0.524 

0.524 

0.525 

0.526 

0.527 

0.529 

0.532 

0.536 

0.539 

0.544 

0.549 

35° 

0.611 

0.611 

0.612 

0.614 

0.617 

0.622 

0.624 

0.630 

0.636 

0.644 

0.653 

40° 

0.698 

0.699 

0.700 

0.703 

0.707 

0.712 

0.718 

0.727 

0.736 

0.748 

0.763 

45° 

0.785 

0.786 

0.789 

0.792 

0.798 

0.804 

0.814 

0.826 

0.839 

0.858 

0.881 

50° 

0.873 

0.874 

0.877 

0.882 

0.889 

0.898 

0.911 

0.928 

0.947 

0.974 

1.011 

55° 

0.960 

0.961 

0.965 

0.972 

0.981 

0.993 

1.010 

1.034 

1.060 

1.099 

1.154 

60° 

1.047 

1.049 

1.054 

1.062 

1.074 

1.090 

1.112 

1.142 

1.178 

1.233 

1.317 

65° 

1.134 

1.137 

1.143 

1.153 

1.168 

1.187 

1.215 

1.254 

1.302 

1.377 

1.506 

70° 

1.222 

1.224 

1.232 

1.244 

1.262 

1.285 

1.320 

1.370 

1.431 

1.534 

1.735 

75° 

1.309 

1.312 

1.321 

1.336 

1.357 

1.385 

1.426 

1.488 

1.566 

1.703 

2.028 

80° 

1.396 

1.400 

1.410 

1.427 

1.452 

1.485 

1.534 

1.608 

1.705 

1.885 

2.436 

85° 

1.484 

1.487 

1.499 

1.519 

1.547 

1.585 

1.643 

1.731 

1.848 

2.077 

3.131 

90° 

1.571 

1.575 

1.588 

1.610 

1.643 

1.686 

1.752 

1.854 

1.993 

2.275 

GO 

METHODS   OF  APPROXIMATE  INTEGRATION       271 

EQc,  4) 


0° 

5° 

10° 

15° 
20° 
25° 

30 

35° 

40° 

45° 
50' 
55 

60° 
65° 

70° 

75° 
80° 
85° 

90° 


£  =  0      A:  =  0.1    £=0.2   ft=0.8 
sinO°      sin  6°     sin  12°    sin  18° 


0.000 
0.087 
0.175 

0.262 
0.349 
0.436 

0.524 
0.611 
0.698 

0.785 
0.873 
0.960 

1.047 
1.134 
1.222 

1.309 
1.396 
1.484 

1.571 


0.000 
0.087 
0.175 

0.262 
0.349 
0.436 

0.523 
0.610 
0.698 

0.785 
0.872 
0.959 

1.046 
1.132 
1.219 

1.306 
1.393 
1.480 


0.000 
0.087 
0.174 

0.262 
0.349 
0.436 

0.523 
0.609 
0.696 

0.782 
0.869 
0.955 

1.041 
1.126 
1.212 

1.297 
1.383 
1.468 


£  =  0.4   £  =  0.5    £=0.6 
sin  24°    sin  80°    sin  87° 


£  =  0.7   £  =  0.8   £=0.9     £=1 
sin  45°    sin  53°    sin  G4°    sin(ju° 


0.000 
0.087 
0.174 

0.262 
0.348 
0.435 

0.521 
0.607 
0.693 

0.779 
0.864 
0.948 

1.032 
1.116 

1.200 

1.283 
1.367 
1.450 


0.000  0.000    0.000 

0.087  0.087    0.087 

0.174  0.174   0.174 

0.261  0.261    0.261 

0.348  0.347    0.347 

0.434  0.433   0.431 

0.520  0.518    0.515 

0.605  0.602    0.598 

0.690  0.685   0.679 

0.773  0.767    0.759 

0.857  0.848    0.837 

0.939  0.928    0.914 


1.566    1.554    1.533 


0.000  0.000  0.000  0.000 

0.087  0.087  0.087  0.087 

0.174  0.174  0.174  0.174 

0.260  0.260  0.259  0.259 

0.346  0.345  0.343  0.342 

0.430  0.428  0.425  0.423 

0.512  0.509  0.505  0.500 

0.593  0.588  0.581  0.574 

0.672  0.664  0.654  0.643 


1.021 
1.103 
1.184 

1.264 
1.344 
1.424 


1.008 
1.086 
1.163 

1.240 
1.316 
1.392 


0.989 
1.063 
1.135 

1.207 
1.277 
1.347 


1.504    1.467    1.417 


0.748 
0.823 
0.895 


0.737 
0.808 

0.875 


0.723  0.707 
0.789  0.766 
0.850   0.819 


0.965   0.940   0.907   0.866 
1.033   1.001   0.960   0.906 


1.099 

1.163 

1.227 
1.289 


1.060 

1.117 
1.172 
1.225 


1.008  0.940 

1.053  0.966 

1.095  0.985 

1.135  0.996 


1.351    1.278   1.173   1.000 


From  Art.  190  we  see  that  a  definite  integral,   I    F(x)dx, 

can  be  interpreted  as  the  area  inclosed  by  the  curve  y  =  F(x), 
the  se-axis,  and  the  ordinates  corresponding  to  the  abscissas  a 

and  b  respectively.     The  value  of  a  definite  integral,  j    F(x)  dx, 

can  therefore  be  found  approximately  by  either  one  of  the 
following  trapezoidal  methods,  or  by  Simpson's  Eule. 

TRAPEZOIDAL  METHODS 

235.  Plot  the  curve  y  =  F(x).  Suppose  that  it  is  as  in 
Fig.  98.  Let  OA  and  OB  represent  the  abscissas  a  and  b 
respectively.  Divide  AB  into  2  n  equal  parts.  Call  each  part 
Ax.     At  A,  B,  and  Al9  A2,  A3,  ••-,  A2n.1}  the  points  of  division 


272       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

of  AB,  erect  ordinates  to  meet  the  curve  in  C,  D,  and  B1}  B2, 
Bs,  •',  Bzn-i  respectively.  Join  CBl9  BXB2,  B2BS,  •  ••,  B2n_1D. 
Denote  the  ordinates  of  the  points  C,  B1}  B2,  B3,  •  ••,  D  by  yly  y2, 

2/3,  -,2/fc+i  respectively. 


Fig.  98. 


By  geometry,  the  area  of  a  trapezoid  is  the  product  of  one 
half  the  sum  of  the  bases  by  the  height.  Therefore  the  sum 
of  the  areas  of  the  trapezoids  ACB^,  A^B^,  A2B2B3A3, 

=  K(yi  +  2/2)  +  (2/2  +  2/3)  +  (2/3  +  2/4)  H h  (2/2»  +  2/2n+i)]  Aa; 

=  Aa?[i  ?/i  +  y2  +  ?/3  +  yA  H h  2/2n  +i2fe«+i]. 

Therefore  an  approximate  value  of  J    F(x)  dx  is 

Aa?  H  2/i  +  2/2  +  2/3  +  2/4  H h  2/2»  +  i  2/2«+i]  • 

In  the  above,  the  number  of  equal  divisions  of  AB  was  taken 
even.     It  could  however  as  well  be  taken  odd. 

In  Fig.  98,  the  curve  was  assumed  to  be  concave  downwards. 
In  such  a  case  the  approximation  is  obviously  less  than  the 
value  of  the  integral.  If  the  curve  is  concave  upwards,  the 
approximation  is  greater  than  the  value  of  the  integral. 

Obviously,  the  greater  the  number  of  parts  into  which  AB  is 
divided  the  closer  will  be  the  approximation. 


METHODS   OF  APPROXIMATE  INTEGRATION      273 

A  second  trapezoidal  formula  may  be  derived  as  follows : 

At  BXi  B3,  •••,  B2n  x  (Fig.  98),  draw  tangents  to  the  curve  to 
meet  the  ord mates  produced. 

The  sum  of  the  areas  of  the  trapezoids  formed  by  the  ordi- 
nates  or  ordinates  produced,  the  cc-axis,  and  the  tangent  lines 

thus  drawn         =  2  Ax  [y2  +  yA  +  y6  -\ \-  y2l J . 

Therefore  an  approximate  value  of   I    F(x)  dx 

•J  a 

=  2  Ax[y2  +  y4  +  y6 -\ h  &»]• 

In  this  case  it  is  necessary  that  the  number  of  divisions  be 
even. 

In  this  case,  if  the  curve  is  concave  downwards,  the  approxi- 
mation is  greater  than  the  integral,  and  if  concave  upwards, 
less. 

As  before,  the  greater  the  number  of  parts  into  which  AB  is 
divided  the  closer  will  be  the  approximation. 

Usuall}7  the  mean  of  (1)  and  (2)  will  give  a  result  closer 
than  either. 

236.  In  illustration  of  the  above  methods,  consider  the  fol- 
lowing example. 

—  if  2ft  =  12. 
1     x 

XT        a        5-1      1  1      .  13  13 

Here  Ax  =  — — -  =  - :   y1  =  -  =  1 ;    y2  =  —  =  -  ;   w„  =  —  =  _  ; 

12  3'  n  1  ,J2  1}  4,J3  If  5' 
_1  J_  _3        _1^  _3        _1  13  1 

9*-2,  yt-21-7'  2/6~2|~8'  y7~3]  V^Si'lO'  V*~^r 
3  _1  13  1       3  _1 

-n;  fco-ji  2/"-4i_-i35  ^-^'ii1  yi3~5* 

— , 
1     cc 

by  the  first  method,     =  i[|  +  f  +  f  +  i  +  f  +  f  +  i  +  A  +  T3T 

=  1.6182, 


274       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

and  by  the  second  method, 

_  2T3     ii_|_3i3_l_l_i    _3_"1 

=  1.5929. 
The  mean  of  these  values  =  1.6055. 

By  integration,  (    —  =  loge  5  =  1.6094. 

c/l     x 

The  error  by  the  first  method  is  therefore  less  than  0.009, 
by  the  second  method,  than  0.007,  and  by  the  taking  the  mean 
of  the  two  results,  than  0.004. 

SIMPSON'S   METHOD  OR  RULE 
237.   This  method  is  based  on  the  following  theorem : 

If  three  points  P,  Q,  and  R  on  the  parabola  whose  equation 
is  in  the  form  y  =  ax2  +  bx  +  c  be  such  that  the  ordinates  of  P 
and  It  are  equidistant  from  Q,  then  the  area  inclosed  by  this 
parabola,  the  cc-axis,  and  the  ordinates  of  P  and  R  is 

|[2,'  +  42/"  +  2/"],  (1) 

where  y',  y",  y'"  are  the  ordinates  of  P,  Q,  and  R  respectively. 

Proof.     Denote  the  abscissas  of  P,  Q,  and  R  by  x'  —  h,  x' 

and  x'  +  h  respectively. 

Jr*x'+h 
(ax2  +  bx  +  c)  dx. 
x'-h 


=  S.[~"(6  x'2  +  2  h-)  +  2  bx'  +  2  c 


(2) 


Since  P,  Q,  and  R  lie  on  the  parabola, 

y'  =  a  (V  -  hf  +  b(x'  —  h)  +  c, 
y"  =  ax'2  +  bx'  +  c, 
y<"  =  a(x'  +  hf  +  b(x'  +  h)  +  C 


METHODS   OF  APPROXIMATE  INTEGRATION      275 
Solve  these  equations  for  a,  b,  and  c. 

b=*l~W"-y')-£i(y'-2y"  +  y"'), 

c=iy"-^'"-y')+^(y'-2y"+y'")- 

Substitute  these  values  for  a,  b,  and  c  in  (2)  and  reduce. 

Therefore  the  required  area  =  -  \jy'  -{-  4?/"  +  y'"],  which  is  the 

o 

value  given  in  (1).     The  theorem  is  therefore  proved. 

238.    Simpson's  Eule  can  now  be  established  as  follows : 

Pass  a  parabola  through  the  points  C,  Bx,  and  B2  (Fig.  99), 
with  its  transverse  axis  parallel  to  the  ?/-axis.     The  equation 


2n-l 


will  be  in  the  form  y  =  ax2  +  bx  +  c,  where  a,  b,  and  c  have 
certain  numerical  values,  a  =£  0.  Therefore,  by  the  above  theo- 
rem, the  area  inclosed  by  this  parabola,  the  a>axis,  and  the 
ordinates  AC  and  A2B2  is 


Ax 

T 


2/i  +  4  y2  +  y5 


In  like  manner,  by  passing  parabolas  through  B2BSB4,  B±B5B6, 
and  so- on  up  till  we  reach  B2n_2B2n_xT>,  we  get  the  set  of  areas 


276       DIFFERENTIAL  AND  INTEGRAL    CALCULUS 

ys  +  iyt  +  yo  I 
y5  +  ±y<>  +  y7  L 


Ax 

Y 

Ax 
3 


Ax 


.Vsn-i  +  4  y2n  +  y2n+i 


Ax 
~3 


Therefore,  by  addition,  the  surn  of  the  areas  of  the  parabolas 
Vi  +  2  (2/3  +  y5  +  y7  H h  y^) 

+  ^  fe  +  2/4  +  2/6  +  •'•  2/2»)  +  2/2n+l 

Therefore  an  approximate  valne  of    I  F(x)dx 
2/i  +  2  (2/3  +  2/5  +  2/7  H h  2/2n-i) 

+  4  (?/2  +  2/4  +  2/6  H 1"  2/2n)  +  2/2n+l 


Simpson's  Rule  is  slightly  more  difficult  of  application  than 
either  of  the  trapezoidal  rules,  but  gives  a  closer  result  than 
either.  It  will  usually  give  a  closer  result  than  that  found  by 
taking  the  mean  of  the  results  found  by  the  trapezoidal  rules. 

When  Simpson's  Rule  is  applied,  the  number  of  divisions  of 
the  line  AB  must  be  even. 


PLANIMETERS 

239.  A  planimeter  is  a  mechanical  contrivance  whereby  a 
plane  area  is  measured  by  passing  a  tracer  around  the  bound- 
ing curves.  One  in  most  common  use  is  Amsler's  Polar  Pla- 
nimeter, a  description  of  which  is  given  in  Carr's  Synopsis  of 
Pure  Mathematics.  There  is  also  a  discussion  on  planimeters 
in  Carpenter's  Text-book  of  Experimental  Engineering. 


METHODS   OF  APPROXIMATE  INTEGRATION       277 
EXERCISES 

X5dx 
— -,  2  n  =  12,  by 

Simpson's  Kuie.  x     Ans.  1.6008. 

For  each  of  the  following  integrals,  for  the  value  of  2  n  set 
opposite  the  integral,  calculate  approximate  values  by  all  rules, 
and  compare  results  with  the  exact  values  of  the  integrals. 

J  10 
log10 x dx,  In  — 10. 

Ans.  First  Trap.  Eule,  6.0656 ;  Second  Trap.  Eule,  6.1374 ; 
Mean,  6.1015 ;  Simpson's  Eule,  6.0896. 

3.  f-^-,  2n  =  6. 

jo   1  4-  x3 

Ans.  First  Trap.  Eule,  1.0885 ;  Second  Trap.  Eule,  1.0946 ; 
Mean,  1.0915;  Simpson's  Eule,  1.0906. 

4.  Cx*dx,  2  ?i  =  12. 

Ans.  First  Trap.  Eule,  320.89 ;  Second  Trap.  Eule,  318.22 ; 
Mean,  319.56 ;  Simpson's  Eule,  320. 


CHAPTER   XXX 

ELEMENTS   0E  KINEMATICS 

For  the  purpose  of  completeness,  those  definitions  and  fun- 
damental principles  of  kinematics  needed  in  the  subsequent 
chapters  are  here  briefly  treated. 

240.  Definition.  The  change  of  position  of  a  particle  from  a 
point  P1  to  a  point  P2  is  called  the  displacement  of  the  particle 
from  Px  to  P2. 

241.  A  displacement  from  a  point  Px  to  a  point  P2  is  known 
when,  and  only  when,  the  magnitude  and  direction  of  the 
straight  line  joining  Py  and  P2  can  be  determined  completely. 
A  displacement  has  therefore  both  length  and  direction. 

The  direction  on  the  line  in  which  displacement  takes  place 
is  indicated  by  an  arrowhead. 

242.  Let  A  and  B  indicate  two  displacements  of  a  particle, 
and  let  Px  be  the  initial  position  of  the  particle.     Draw  PXP2 

equal  to  A  and  in  the  same  direction 

*-  with  it.     From  P2  draw  P2P3  equal  to 

B  and  in  the  same  direction  with  it. 

The  two  displacements  PiP2  and 
P2P3  are  equivalent  to  the  single  dis- 
placement PiP3. 

Complete  the  parallelogram  PXP2P^P4. 

P*  p3  Since   P1P2  =  P4P3,   and   P»P3  =  P1P4, 

Fig.  100.  ,  _.     n  _,  _  n    _.  _^ 

the  two  displacements  PXPA  and  P4P3, 

or  B  and  A,  are  equivalent  to  the  single  displacement  PXP3. 

In  the  case  of  two  displacements,  therefore,  the  order  in  which 

the  displacements  are  made  is  immaterial. 

278 


ELEMENTS   OF  KINEMATICS  279 

To  find  the  single  displacement  which  is  equivalent  to  two 
given  displacements,  we  lay  off  the  two  given  displacements 
as  two  sides  of  a  triangle  with  the  arrowheads  the  same  way 
round.  The  remaining  side  of  the  triangle  with  the  arrow- 
head reversed  is  the  single  displacement  equivalent  to  the 
two  given  displacements. 

243.  Let  A,  B,  and  C  be  three  displacements,  and  Pv  the 
initial  position  of  the  particle.  From  P1  draw  PXP2  equal  to  A 
and  in  the  same  direction  with  it.  From  P2  draw  P2PS  equal 
to  B  and  in  the  same  direction  with  it.  From  P3  draw  P^P4 
equal  to  C  and  in  the  same  direction  with  it. 

The  displacements  PXP2  and  P2P3  are  equivalent  to  the  dis- 
placement PiP3,  and  the  displacements  PiP3  and  P3P4  to  the  dis- 
placement PiP±.  Then  the  displacements  PXP2,  P2P&  and  P3P4, 
or  A,  B,  and  C,  are  equivalent  to  the  single  displacement  PXPA. 

Also,  as  in  two  displacements,  it  can  be  readily  shown  that 
the  order  in  which  the  displacements  are  made  is  immaterial. 

To  find  the  single  displacement  which  is  equivalent  to  the 
three  given  displacements,  we  lay  off  the  three  given  displace- 
ments as  the  sides  of  a  quadrilateral  with  the  arrowheads  the 
same  way  round.  The  remaining  side  of  the  quadrilateral  with 
the  arrowhead  reversed  is  the  single  displacement  equivalent 
to  the  three  given  displacements. 

244.  A  displacement  can  be  resolved  into  any  number  of 
component  displacements  parallel  to  given  lines. 

Let  PXP2  be  a  given  displacement.     Let  us,  for  defmiteness, 
resolve  it  into  two  displacements  paral- 
lel to  the  lines  A  and  B. 

From  Px  draw  a  line  parallel  to  A, 
and  from  P2  draw  a  line  parallel  to  B. 
Let  P3  be  the  point  of  intersection  of 
these  lines. 

Then  PiP2  is  resolved  into  two  dis-  ^" 

placements  parallel  to  A  and  B.  Fig.  101. 


280       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

245.  Definition.  The  mean  velocity  of  a  moving  particle 
during  a  given  time  is  a  quantity  which  has  for  direction  the 

,p  direction  of  the  displacement  of  the  particle 
in  that  time,  and  for  magnitude  the  magni- 
tude of  the  displacement  divided  by  the 
time. 

Thus,  if   a   particle   move   on   any  path 
Fig.  102.  from  P1  to  P2  in  a  given  time  t,  its  mean 

velocity  has  the  direction  of  the  straight  line  PiP2?  an(i  the 
magnitude  the  length  of  the  line  divided  by  t. 

Mean  velocity  must  be  distinguished  from  mean  speed. 
The  mean  speed  of  the  particle  in  the  above  illustration  is 

arc  PiP2. 

t 

The  mean  velocity  of  a  particle  in  general  varies  with  the 
time.  If  it  does  not,  both  the  direction  of  the  particle  and  the 
quotient  of  the  magnitude  of  its  displacement  by  the  time 
must  be  constant.  In  such  a  case  the  particle  moves  in  a 
straight  line  and  with  uniform  speed. 

246.  Definitions.  A  particle  which  moves  in  a  straight  line 
with  uniform  speed  is  said  to  have  uniform  velocity. 

The  velocity  of  a  particle  at  a  given  instant  is  the  limit 
which  the  mean  velocity  of  the  particle  for  a  period  of  time 
immediately  succeeding  the  instant  in  question  approaches  as 
the  period  of  time  is  allowed  to  decrease  without  limit. 

247.  Suppose  that  a  particle  moving  on  a  known  path  passes 
over  a  distance  s  in  a  time  t  To  find  the  magnitude  and  direc- 
tion of  the  particle  at  any  given  instant. 

Let  Px  be  the  position  of  the  particle  at 
the  instant  in  question,  and  P2  its  position 
at  the  end  of  a  time  t. 

Denote  the  chord  PXP2  by  Ac,  and  the 
arc  PXP2  by  As.     Since  s  increases  as  t  increases,  both  Ac  and 
As  are  positive. 


Fig.  103 


ELEMENTS   OF  KINEMATICS 


281 


Ac 
By  definition,  —  =  average  velocity  of  the  particle   for   a 

time  At  immediately  succeeding  the  instant  in  question,  and 

=  velocity,  v,  of  the  particle  at  the  instant  in  question. 


limit 
At  =  0 


Ac 

At 


Now 


limit 
At  =  0 


"Ac" 
A* 


.-.  v  = 


limit 


limit 
At±0 

limit 
At=0 

ds 
dt' 

ds 
di 


"As 

_At 

"As" 
At 

"As" 

At 


Ac" 
As 


limit 
At  =  0 


Since 


"Ac 
_As_ 

limit 
At=0 


"Ac" 
As 


1, 


Therefore  the  velocity,  v,  of  the  particle  at  any  given  instant 

ds 

=  — ,  where  t  has  the  value  corresponding  to  the  instant  in 

question. 

As  At  =  0,  the  chord  PYP2  approaches  as  its  limit  the  tangent 
line  to  the  curve  at  P±.  Therefore  the  direction  of  the  velocity 
at  a  given  instant  is  that  of  the  tangent  line  to  the  curve  at  the 
point  which  the  particle  occupies  at  that  instant. 

248.  In  Chapter  III,  and  the  preceding  article,  we  have  sup- 
posed that  s  is  the  distance  passed  over  by  a  particle  in  a  time  t. 
We  shall  find  it  convenient,  however,  to  take  s  sometimes  not  as 
the  distance  passed  over  by  the  particle  in  a  given  time  but  as 

a  constant,  minus  this  distance.  , 

■  i 1  i 

Thus,  if  the  particle  moves  in  a         ^         ^ 
time  t  from  a  point  A  to  a  point  P 

under  a  force  directed  towards  a  fixed  point  0  and  varying 
as  some  power  of  the  distance  of  the  particle  from  0,  it  will 
be  convenient,  as  we  shall  see  later,  to  take  OP,  not  AP,  equal 
to  s.     Then  ^P=0^1-s. 


282       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

When  the  distance  passed  over  by  a  particle  in  a  given  time 

is  taken  as  a  constant,  minus  s,  then  v  = •     In  this  case  s 

dt 
decreases  as  t  increases. 

249.  The  magnitude  of  a  velocity  is  the  magnitude  of  the 
displacement  during  the  unit  of  time,  or  the  magnitude  of  the 
displacement  if  the  particle  moved  with  uniform  velocity  during 
the  unit  of  time.  The  direction  of  a  velocity  is  the  direction 
of  the  displacement  in  the  unit  of  time,  or  the  limit  which  this 
displacement  for  a  period  of  time  approaches  as  the  period  of 
time  is  allowed  to  decrease  without  limit.  Then,  since  velocity 
has  a  definite  magnitude  and  a  definite  direction,  it  follows 
that  velocities  may  be  compounded  or  resolved  into  component 
velocities  in  the  same  manner  as  displacements. 

250.  When  the  velocities  of  a  particle  at  two  points  are 
known,  the  change  in  velocity  can  be  found  as  follows : 


Fig.  105. 

Let  v1  represent  in  magnitude  and  direction  the  velocity  at  P1? 
and  v2  the  velocity  at  P2  (Fig.  105).  At  any  point  A  (Fig.  106), 
draw  AB  and  AC  equal  and  parallel  to  i\  and  v2  respectively, 
with  the  arrowheads  towards  B  and  C. 

The  sum  of  v1  and  BC  is  v2.  Therefore  BC  represents  in 
magnitude  and  direction  the  change  in  velocity  of  the  particle 
between  P1  and  P2. 

Change  in  velocity  must  be  distinguished  from  change  in 
speed. 

Thus,  the  change  in  speed  in  the  above  illustration  is  the 
length  of  v2,  minus  the  length  of  vx. 


ELEMENTS   OF  KINEMATICS  283 

251.  Definitions.  The  mean  acceleration  of  a  moving  particle 
in  a  given  time  is  a  quantity  which  has  for  direction  the  direc- 
tion of  the  change  in  velocity  during  that  time,  and  for  magni- 
tude the  magnitude  of  the  change  in  velocity  divided  by  the 
time. 

Thus,  in  the  illustration  of  Art.  250,  if  t  is  the  given  time, 

is  the  mean  acceleration. 

t 

The  acceleration  of  a  particle  at  a  given  instant  is  the  limit 
which  the  mean  acceleration  of  the  particle  for  a  period  of 
time  immediately  succeeding  the  instant  in  question  approaches 
as  the  period  of  time  is  allowed  to  decrease  without  limit. 

If  a  particle  moves  in  such  a  manner  that  the  acceleration 
at  every  instant  is  the  same  in  magnitude  and  direction,  it  is 
said  to  move  with  uniform  acceleration. 

252.  The  magnitude  of  an  acceleration  is  the  magnitude  of 
the  change  in  velocity  during  the  unit  of  time,  or  the  magni- 
tude of  the  change  in  velocity  if  the  particle  moved  with 
uniform  acceleration  during  the  time.  The  direction  of  an 
acceleration  is  the  direction  of  the  change  in  velocity  during 
the  unit  of  time,  or  the  limit  which  the  change  in  velocity 
approaches  as  the  period  of  time  decreases  without  limit. 
Then,  since  acceleration  has  a  definite  magnitude  and  a  defi- 
nite direction,  it  follows  that  accelerations  may  be  compounded 
or  resolved  into  component  accelerations  in  the  same  manner 
as  velocities  or  displacements. 

253.  To  find  the  magnitude  and  direction  of  an  acceleration 
at  any  instant  in  terms  of  the  magnitude  and  direction  of  the 
velocity  at  that  instant : 

Let  Pj  be  the  position  of  the  particle  at  the  instant  in  ques- 
tion (see  Fig.  107).  Let  P2  be  the  position  of  the  particle 
after  a  time  At.  Let  aN  be  the  acceleration  along  the  normal 
line  and  aT  the  acceleration  along  the  tangent  line,  at  Pv 


284       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


Let  P^  and  P2R2  represent  in  magnitude  and  direction  the 
velocities  v  and  v  +  Av  at  P1  and  P2  respectively.  At  Px  draw- 
Pi^  equal  and  parallel  to  P2R2. 
Join  R^.  From  JS  drop  per- 
pendiculars on  the  normal  and 
tangent  lines  at  P1  to  meet  these 
lines  in  Q  and  T  respectively. 

The  line  R^S  represents  in 
magnitude  and  direction  the 
change  in  velocity,  Av,  during 
the  time  At. 

Denote  the  arc  P^P2  by  As. 
At   first  suppose  that   s   in- 
creases as  t  increases,  so  that 
As  is  positive. 

The  projection  PXQ  of  RXS  on  the  normal  line  is  the  change 
in  velocity  along  the  normal  line  during  the  time  At     Then 


Fig.  107. 


limit 
A£=0 


At 


is  the  acceleration  along  the  normal  line  at  P1}  or  aN. 


Denote  the  angle  between  P^RX  and  P2R2  by  A<£. 


Then  a. 


limit 
At  =  0 

-  At  _ 

limit 

At=0 

limit 

At=0 

P^sinAcfr 

At 


',     .    A   xsmA<A     Ad>     As 

(v  +  Av) •  — -  •  — - 

v  J    A</>         As     Af 


P 

v2 


where  p  is  the  radius  of  curvature  of  the  curve  at  Px. 

The  projection  R^T  of  RXS  on  the  tangent  line  is  the  change 
in  velocity  along  the  tangent  line  during  the  time  A*.     Then 


limit 
At  =  0 

or  aT. 


RXT 

At 


is  the  acceleration  along  the  tangent  line  at  PD 


ELEMENTS   OF  KINEMATICS 


285 


Then 

limit 

Ut~  At  =  0 


At 


limit 
At=0 


limit 
At  =  0 


limit 

A£  =  0 


limit 

At~0 


limit 
At  =  Q 


i\s  cos  Acf>  -  pji; 

At 

(v  +  A?;)  cos  Ad>  —  v 

At 


v  (cos  A<fr  —  1)  +  Av  cos  Aj>" 

~~aT 


Z  v  sin 


2i^ 


At 
"sin  -|  Ad> 


,  Av         A  , 
H cos  Ad> 

A£ 


^    n  dd>  ,  dv 
=  1  •  ()•— - *-  H 


,  .   .    Ad>  .  Av        .    ' 
v  sin  A  Ad>  •  — - "  4-  —  cos  Ac/> 

2     ^    A*       At 


dv 
~dt 


d2s 
dt2' 


The  resultant  acceleration  is  therefore 


=  V«/  +  aT2 


lv*  ,  (dv\ 


and  has  the  direction  0  =  tan-1  — — ,  where  0  is  the  angle  be- 

dv  ° 

Pdt 
tween  the  tangent  to  the  curve  at  the  point  Px  and  the  direction 
of  the  resultant  acceleration. 

If  s  decreases  as  t  increases,  aN  is  as  before,  while  aT— 

2       '  dt2 

v 
If  motion  is  along  a  straight  line,   — ,  or  aV7  is  zero.     Then 

P 

in  motion  along  a  straight  line,  the  magnitude  of  the  accelera- 
tes           d~s 
tion  a  is  —  or ,  the  +  or  —  sign  being  taken  according 

as  s  increases  or  decreases  as  t  increases,  and  the  direction  of 
the  acceleration  is  in  the  direction  of  the  motion. 


/ 


286       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

EXERCISES 

1.  Show  that  the  resultant  P\Pn  of  any  number  of  compo- 
nent displacements  PXP2,  PiP®  •••?  Pn-\Pn  is  the  diagonal  of  a 
rectangle  PxAPnB  in  which  PXA  and  PXB  are  the  sums  of  the 
displacements  resolved  along  the  lines  PXA  and  PXB  respec- 
tively, which  are  perpendicular  to  each  other. 

2.  A  body  undergoes  three  component  displacements,  30 
feet  K,  60°  E. ;  40  feet  S. ;  50  feet  W.,  30°  K  Find  the  result- 
ant displacement.  Ans.  10 V3  ft.  W. 

3.  A  wheel  4  feet  in  diameter  rolls  along  a  horizontal  road, 
turning  through  an  angle  of  30°.  Find  the  displacement  of  the 
point  of  the  wheel  initially  in  contact  with  the  road  relative  to 
the  point  on  the  road  with  which  it  was  in  contact. 

Ans.  0.272  ft.,  inclined  80°  (approx.)  to  the  horizontal. 

4.  A  ship  is  carried  by  her  screw  12  miles  N".,  by  the  wind 
4  miles  E.,  15°  K,  and  by  the  current  2  miles  S.,  15°  E.  Find 
the  component  in  a  northeasterly  direction  of  the  resultant 
displacement.  Ans.  10.95  mi. 

5.  A  ship  is  influenced  by  her  screw,  the  wind,  and  a  cur- 
rent. Her  screw  carries  her  12  miles  N.,  the  wind  carries  her 
3  miles  S.,  15°  E.,  and  it  is  observed  that  her  total  displace- 
ment is  15  miles  in  a  northeasterly  direction.  Find  her  dis- 
placement due  to  the  current  and  the  direction  of  the  current. 

Ans.  9.94  mi.  E.,  8°  42'  K 

6.  A  particle  moves  100  feet  W.,  30°  K  in  30  seconds,  and 
thenGe  50  feet  S.,  60°  W.  in  20  seconds.  Find  (a)  the  mean 
speed ;  (6)  the  mean  velocity  of  the  particle  during  its  time 
of  motion. 

Ans.  (a)  3  ft.  per  sec. ;  (6)  2.645  ft.  per  sec.  W.,  10°  53.6'  N. 

7.  An  automobile  is  running  north  at  the  rate  of  16  miles  per 
hour.  A  carriage  going  at  the  rate  of  8  miles  per  hour  appears 
to  a  person  in  the  automobile  to  be  going  at  the  rate  of  20  miles 
per  hour.     Find  the  direction  of  the  carriage. 

Ans.  71°  47.4'  W.  or  E.  of  S. 


ELEMENTS   OF  KINEMATICS  287 

8.  A  train  is  running  at  the  rate  of  50  miles  per  hour  in  a 
shower  of  rain.  If  the  drops  fall  vertically  with  a  speed  of 
200  feet  per  second,  in  what  direction  will  they  seem  to  a  man 
on  the  train  to  fall  ? 

Ans.  At  an  angle  of  20°  8'. 2  with  the  vertical. 

9.  A  river  has  a  current  which  runs  at  the  rate  of  a  miles 

per  hour.     A  steamer  whose  speed  is  b  miles  per  hour  in  still 

water  is  to  be  run  straight  across.     Find  the  direction  in  which 

she  must  be  steered.  Ans.  At  cos"1-  with  the  river  bank. 

b 

10.  A  train  running  at  the  rate  of  50  miles  per  hour  is  hit 
by  a  stone  moving  horizontally  and  at  right  angles  to  the  track, 
at  the  rate  of  30  feet  per  second.  Find  the  magnitude  of  the 
velocity  with  which  the  stone  hits  the  train,  and  the  angle 
which  it  makes  with  the  direction  of  motion  of  the  train. 

Ans.  79.2  ft.  per  sec;  tan-1^. 

11.  The  initial  and  final  velocities  of  a  body  which  moves 
for  ,3  hours  is  10  miles  per  hour  E.,  30°  N.,  and  5  miles  per  hour 
N.  respectively.     Find  the  mean  acceleration  of  the  body. 

Ans.  -§V3  mi.-per-hr.  per  hr.  W. 

12.  A  particle  moves  with  uniform  speed  along  the  circum- 
ference of  a  circle  of  radius  6  feet  from  one  end  of  a  diameter 
to  the  other  in  10  seconds.  Find  (a)  the  mean  speed, 
(b)  the  mean  velocity,  of  the  particle  for  the  time  of  motion. 

Ans.  1.88  ft.  per  sec. ;  (b)  1.2  ft.  per  sec.  in  the  direction 
of  the  diameter. 

13.  In  Exercise  12,  show  that  the  acceleration  is  directed 
towards  the  center  of  the  circle  and  is  0.592  ft.-per-sec.  per  sec. 

14.  A  horse  travels  with  uniform  speed  on  an  elliptical  race 
track,  semi-axes  800  and  400  feet  respectively,  from  one  end  of 
the  major  axis  to  the  other  in  1  minute  25  seconds.  Find  (a) 
the  mean  speed ;  (b)  the  mean  velocity  of  the  horse  during  the 
time  of  motion.         Ans.  15.52  mi.  per  hr. ;  12.83  mi.  per  hr.  in 

the  direction  of  the  major  axis. 


CHAPTER   XXXI 
FOKCE.    MASS.    EQUATION   OF  MOTION   OF  A  PAETIOLE 

254.  Our  idea  of  force  is  ultimate,  like  that  of  color,  taste, 
and  smell,  and  like  them  cannot  be  described.  We  all  have 
the  idea  of  force,  however.  When  one  speaks  of  exerting  force 
we  know  exactly  what  he  means. 

255.  Let  us  first  choose  as  the  unit  of  force  that  force  exerted 
by  a  given  spring  when  stretched  by  a  given  amount.  Suppose 
that  we  have  several  exactly  similar  springs.  Then  if  they 
are  stretched  by  the  same  amount,  they  exert  the  same  forces. 
Now  suppose  that  we  attach  one  of  these  springs  to  a  body 
lying  on  a  smooth  table,  and  keeping  it  always  stretched  by 
the  given  amount  and  in  such  a  position  that  the  force  exerted 
by  it  acts  in  a  horizontal  straight  line,  allow  it  to  act  on  the 
body  during  known  periods  of  time.  By  noting  the  distances 
through  which  the  body  moves,  we  are  able  to  determine  the 
magnitude  and  direction  of  the  acceleration  produced.  The 
same  determinations  may  be  made  with  two  or  more  springs. 
These  experiments,  although  of  necessity  rough,  will  never- 
theless be  sufficient  to  suggest  the  following  fundamental 
principles  of  motion  :  — 

(1)  In  all  cases  the  acceleration  is  uniform. 

(2)  In  all  cases  the  direction  of  the  acceleration  is  that  of 
the  force. 

(3)  The  acceleration  produced  by  the  force  is  proportional 
to  the  force,  double  the  force  producing  double  the  acceleration, 
three  times  the  force  three  times  the  acceleration,  and  so  on. 

(4)  The  accelerations  do  not  depend  on  the  lengths  of  time 

during  which  the  forces  act. 

288 


FORCE.      MASS  289 

256.  The  first  of  the  results  of  the  preceding  article  shows 
that  the  acceleration  is  the  same  whether  the  body  starts  from 
rest  or  with  an  initial  velocity.  The  third  shows  that  if  the 
force  is  zero,  the  acceleration  is  zero,  and  that  therefore  the 
body,  if  initially  at  rest,  remains  at  rest,  and  if  initially  in 
motion,  remains  in  motion  with  a  velocity  equal  to  the  initial 
velocity. 

257.  If  we  indicate  the  magnitudes  of  the  force  and  accel- 
eration by  F  and  a  respectively,  the  third  of  the  results  of 

Art.  255  gives  Faza. 

F  . 
If  Fcca,  —  is  constant. 
a 

258.  Definition.     We  shall  define  the  mass  of  the  body  by 

F 

saying  that  it  is  a  number  proportional  to  the  constant  ratio  — • 

F  a 

Thus,  if  m  denotes  the  mass  of  a  body,  moo  — 

F  F  a      ,.  , 

If  —  ocm,  —  =  km,  where  A:  is  a  constant  depending  on  the 
a  a 

units  chosen  for  F,  m,  and  a. 

259.  So  far  we  have  supposed  that  the  forces  remain  con- 
stant both  in  magnitude  and  direction  during  the  whole  time 
of  the  motion.  Let  us  now  suppose  that  they  vary  in  magni- 
tude or  direction,  or  in  both  magnitude  and  direction.  Sup- 
pose that  at  a  particular  time  t  the  force  acting  is  F,  and  the 
acceleration  produced  by  the  force  is  a.  If  F  remained  con- 
stant in  magnitude  and  direction  for  a  period  of  time  M 
immediately  succeeding  t0,  the  equation  F—kma  would  hold 
true  for  that  time.  Then,  since  the  equation  is  independent 
of  the  time  of  motion,  it  would  hold  true  if  At  be  taken  as 
small  as  we  please.  From  this  we  conclude  that  the  limit, 
as  At  approaches  zero,  of  F—kma  is  F=kma.  That  is,  the 
variable  force  F  at  any  time  t0  is  such  that  F=kma,  where  a 
is  the  acceleration  of  the  particle  at  the  time  t0,  and  has  the 
direction  of  the  force  at  that  time. 


290       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Definition.  The  equation  F=kma  is  called  the  equation  of 
motion  of  a  particle. 

260.  Let  F  and  F'  be  two  forces  which,  acting  on  a  body  of 
mass  m,  produce  accelerations  of  a  and  a'  respectively. 

Then  F  =  kma, 

and  F'  =  kma'. 

.-.  F:F'::a:a'. 

Then  to  compare  two  forces  we  may  allow  them  to  act  on 
the  same  body  and  compare  the  two  accelerations. 

Let  F  be  a  force  which,  acting  on  two  bodies  of  masses  m 
and  m',  produces  the  accelerations  a  and  a'  respectively. 

Then  F—  kma, 

and  F—km'd. 

.*.  m  :  m' : :  a' :  a. 

Then  to  compare  the  masses  of  two  bodies  we  may  allow  the 
same  force  to  act  on  the  bodies  and  compare  the  accelerations. 

261.  Definition.  The  force  which  the  earth  exerts  on  a  body 
in  the  neighborhood  of  the  earth  is  called  the  weight  of  the 
body. 

262.  We  saw  in  Art.  260  how  force  and  mass  can  be 
measured.     It  remains  to  choose  units  for  them. 

It  is  desirable  to  choose  units  for  them  so  that  k  will  have 
the  value  1.  We  shall,  therefore,  assign  a  unit  arbitrarily  to 
either  the  force  or  mass,  and  define  the  other  unit  so  that  unit 
of  force,  acting  on  unit  of  mass,  produces  unit  of  acceleration. 

263.  There  are  two  systems  of  units  in  common  use :  the 
Gravitational  System  and  the  Absolute  System. 

In  the  Gravitational  System,  the  unit  of  force  is  chosen  arbi- 
trarily and  the  unit  of  mass  is  defined  to  be  that  mass  which 
acted  upon  by  unit  of  force  produces  unit  of  acceleration. 


EQUATION   OF  MOTION   OF  A   P ARTICLE  291 

In  the  Absolute  System,  the  unit  of  mass  is  chosen  arbitrarily, 
and  the  unit  of  force  is  defined  to  be  that  force  which  acting 
on  unit  mass  produces  unit  of  acceleration. 

264.  In  the  Gravitational  System  of  units,  we  may  use  the 
foot-pound-second  (F.P.S.)  System,  or  the  centimeter-gram- 
second  (C.Gr.S.)  System. 

In  the  F.P.S.  Gravitational  System,  the  unit  of  force  is 
defined  to  be  the  weight  of  a  certain  piece  of  platinum,  called 
the  pound,  stored  in  the  Standards  Office  in  London.  The 
unit  of  acceleration  is  1  foot-per-second  per  second.  To  de- 
termine the  unit  of  mass :  The  weight  of  the  pound  produces 
in  its  mass  an  acceleration  of  g  feet-per-second  per  second. 
Then,  since  the  masses  are  inversely  proportional  to  the  accel- 
erations (see  Art.  260),  the  unit  of  mass  must  be  g  times  the 
mass  of  the  pound. 

In  this  system  of  units,  therefore, 

F  =  ma, 

where  F  indicates  the  force  expressed  in  pounds  weight,  m 
the  mass  expressed  in  terms  of  a  unit  which  is  g  times  the 
mass  of  the  pound,  and  a  the  acceleration  in  feet-per-second 
per  second. 

In  the  C.G.S.  Gravitational  System,  the  unit  of  force  is  the 
weight  of  the  one-thousandth  part  of  a  piece  of  platinum  called 
the  kilogram  stored  in  the  Palais  des  Archives  in  Paris.  The 
unit  of  acceleration  is  the  centimeter-per-second  per  second. 
To  determine  the  unit  of  mass:  The  weight  of  the  gram 
produces  in  a  gram  mass  an  acceleration  of  g'  centimeters- 
per-second  per  second,  where  g'  =  g  times  the  length  of  the  foot 
in  centimeters.  Then  the  unit  of  mass  must  be  g'  times  the 
mass  of  the  gram. 

In  this  system  of  units,  therefore, 

F  =  ma, 
where  F  indicates  the  force  expressed  in  grams  weight,  m 


292       DIFFERENTIAL  AND  INTEGRAL    CALCULUS 

the  mass  expressed  in  terms  of  a  unit  which  is  g'  times  the 
mass  of  the  gram,  and  a  the  acceleration  in  centiineters-per- 
second  per  second. 

265.  In  the  F.P.S.  Absolute  System,  the  unit  of  mass  is 
defined  to  be  the  mass  of  the  pound.  The  unit  of  force  will 
therefore  be  that  force  which  acting  on  the  unit  of  mass  pro- 
duces an  acceleration  of  1  foot-per-second  per  second.  The 
weight  of  the  pound  produces  in  the  unit  of  mass  an  accelera- 
tion of  g  feet-per-second  per  second.  Then,  since  forces  are 
proportional  to  the  accelerations,  the  force  that  produces  the 
acceleration  of  1  foot-per-second  per  second  in  the  mass  of  the 

pound  is  — 
9 

Definition.     The   force  that   produces   in   the   mass   of  the 

pound  an  acceleration  of  1  foot-per-second  per  second  is  called 

the  poundal. 

The  poundal  is  -  times  the  weight  of  the  pound,  or  the  weight 
9 
of  about  half  an  ounce. 

In  this  system  of  units,  therefore, 

F=  ma, 

where  F  indicates  the  force  in  poundals,  m  the  mass  in  pounds, 
and  a  the  acceleration  in  feet-per-second  per  second. 

In  the  C.G.S.  Absolute  System,  the  unit  of  mass  is  the  mass 
of  the  one-thousandth  part  of  the  kilogram.  The  unit  of 
force  is  the  force  which  produces  in  the  unit  of  mass  an  accel- 
eration of  1  centimeter-per-second  per  second.     This  force  is 

—  times  the  weight  of  the  gram. 
9' 

Definition.  *  The  force  which  produces  in  the  mass  of  the 

gram  an  acceleration  of  1   centimeter-per-second  per  second 

is  called  the  dyne. 

The  dyne  is  —  times  the  weight  of  the  gram,  or  the  weight 
9' 
of  -g-iy  of  a  gram  (nearly) 


EQUATION  OF  MOTION  OF  A  PARTICLE  293 

In  this  system  of  units,  therefore, 

F  =  ma, 

where  F  indicates  the  force  in  dynes,  m  the  mass  in  grams, 
and  a  the  acceleration  in  centimeters-per-second  per  second. 

266.  The  numerical  relation  between  the  Gravitational  and 
the  Absolute  systems  of  units  are : 

The  weight  of  1  gram   =  g'  dynes. 

The  weight  of  1  pound  =  g  poundals. 

267.  Definition.  The  product  of  the  mass  of  a  body  and  its 
velocity  in  any  given  direction  is  called  the  momentum  of  the 
body  in  that  direction.  It  is  also  sometimes  called  the  quantity 
of  motion  of  the  body,  or  merely,  the  motion  of  the  body. 

Since  the  momentum  is  mv,  the  rate  of  change  of  momentum 
is  ma. 

268.  The  results  of  a  physical  experiment  are  at  best  more 
or  less  inaccurate,  and  consequently  a  law  derived  from  experi- 
ments should  be  subjected  to  all  possible  tests  before  being 
accepted  as  true.  The  experiments  mentioned  in  Art.  255  as 
leading  to  the  law  of  motion  F=ma  are  so  extremely  rough 
as  to  be  of  no  value  in  the  determination  of  this  law.  They 
merely  serve  to  suggest  that  such  a  law  may  be  possible. 
Then,  before  this  law  is  accepted  as  the  principle  governing 
motion,  it  should  be  subjected  to  further  tests.  Such  tests 
have  been  made  in  a  number  of  ways,  and  in  all  cases  the  law 
has  been  found  satisfactory.  One  of  the  most  convincing  tests 
is  that  made  continually  by  astronomers,  when,  by  assuming 
this  law,  they  find,  by  direct  observation  and  mathematical 
reasoning,  the  exact  positions  which  heavenly  bodies  will 
occupy  at  certain  times. 

269.  Without  entering  further  into  the  discussion,  we  shall 
assume  the  truth  of  the  law  that  F  =  ma  and  shall  also  assume 


294       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

as  true  the  law  first  enunciated  by  Galileo,  and  therefore  bear- 
ing his  name,  which  is  expressed  by  Thompson  and  Tait  in  the 
following  words : 1 

"When  any  forces  whatever  act  on  a  body,  then,  whether 
the  body  be  originally  at  rest  or  moving  with  any  velocity  in 
any  direction,  each  force  produces  in  the  body  the  exact  change 
of  motion  which  it  would  have  produced  if  it  had  acted  singly 
on  the  body  originally  at  rest." 

270.  Since,  by  Galileo's  law,  each  of  several  forces  acting  on 
a  body  produces  the  same  momentum  and  therefore  the  same 
rate  of  change  of  momentum  as  if  it  acted  singly  on  the  body, 
it  follows  that  forces  may  be  compounded  or  resolved  into 
components  in  the  same  manner  as  accelerations  and  therefore 
as  velocities  or  displacements. 

EXERCISES 

In  the  exercises  of  this  chapter  and  the  following  chapters, 
assume  that 

1  pound  avoirdupois  =  453.59  grams. 

1  meter  =  3.284  feet. 

g  =  32. 

1.  Two  forces  produce  in  two  masses,  accelerations  of  20  and 
25  units  respectively.  Find  (a)  the  ratio  of  the  forces  if  the 
masses  are  equal ;  (b)  the  ratio  of  the  masses  if  the  forces  are 
equal.  .  Ans.   (a)  4:5;  (b)  5  :  4. 

2.  Compare  the  values  for  the  mass  of  a  body  expressed  in 
the  ft.-pound-sec.  and  the  yd.-ton-min.  gravitational  systems. 

Am.  24  x  405 :  4. 

3.  Compare  the  values  for  the  mass  of  a  body  expressed  in 
the  cm.-gr.-sec.  and  the  meter-kilogram-min.  systems. 

Ans.  36  X  403 :  4. 

1  See  MacGregor,  Kinematics  and  Dynamics,  p.  204. 


EQUATION   OF  MOTION   OF  A   PARTICLE  295 

4.  Convert  1  poundal  to  absolute  units  in  the  yd.-ton-inin. 
system.  Ans.  f. 

5.  Convert  1  poundal  to  dynes.  Ans.  13,824.7. 

6.  Convert  1  pound  to  dynes.  Ans.  442,392. 

7.  Find  the  acceleration  produced  in  a  mass  of  10  pounds  by 
a  force  of  20  dynes. 

Ans.  0.000145  ft.-per-sec.  per  sec.  (approx.). 

8.  Find  the  force  expressed  in  pounds  which  produces  in  a 
mass  of  20  pounds  an  acceleration  of  12  yd.-per-min.  per  second. 

Ans.    — 
9 

9.  Find  the  mass  on  which  a  force  of  12  pounds  produces  an 
acceleration  of  20  ft.-per-sec.  per  sec.  *        3j/  y, 

5 

10.  Find  the  resultant  of  two  forces  of  3  and  5  pounds  respec- 
tively, acting  on  a  particle  at  an  angle  of  60°. 

Ans.  The  resultant  is  a  force  of  7  lb.  acting  at  an  angle  of 
about  21°  47'  with  the  force  of  5  lb. 

11.  Forces  of  2,  3,  and  5  pounds  respectively  acting  on  a  par- 
ticle and  in  one  plane  make  angles  of  120°  with  one  another. 
Find  the  resultant  of  the  forces. 

Ans.  The  resultant  is  a  force  of  V7  lb.  acting  at  an  angle  of 
about  19°  6'  with  the  force  of  5  lb.  and  about  40°  54' 
with  the  negative  direction  of  the  force  of  2  lb. 

12.  Three  equal  forces,  P,  acting  on  a  particle  and  in  one 
plane  make  angles  of  120°  with  one  another.  Find  the  result- 
ant of  the  forces.  Ans.  The  resultant  is  zero. 

13.  A  man  weighs  160  pounds.  Find  his  apparent  weight 
when  in  an  elevator  which  descends  with  an  acceleration  of 
2  ft.-per-sec.  per  sec.  Ans.  150  lb. 

14.  A  man  who  weighs  160  pounds  notices  that  in  an  elevator 
his  apparent  weight  is  180  pounds.  Show  that  the  elevator 
has  an  upward  acceleration  of  4  ft.-per-sec.  per  sec. 


CHAPTER  XXXII 

EEOTILINEAR  MOTION 

In  this  chapter,  we  shall  consider  the  force  or  forces  acting 
on  the  particle  to  be  such  that  motion  takes  place  in  a  straight 
line. 

MOTION   UNDER  A  CONSTANT  FORCE 

271.  The  simplest  case  of  motion  is  that  of  a  particle  mov- 
ing under  a  constant  force. 

Suppose  that  a  particle  of  mass  m  is  acted  upon  by  a  constant 
force  /  in  the  direction  of  its  motion.  To  determine  the  motion 
of  the  particle. 

Take  0  (Fig.  108),  the  initial  position  of  the  particle,  as  the 
origin,  and  the  direction  of  motion  as  the  positive  direction  of 

the  axis. 

i 1 

O  s  Let  s  be  the  abscissa  of  the  par- 

Fig.  108.  .    " 

tide  at  any  time  t. 

We  shall  suppose  thafc  m,  /,  and  a  are  expressed  in  terms  of 

the  unit  of  mass,  force,  and  acceleration  respectively,  in  some 

chosen  system  of  units. 

Then  ma  =  f. 

d  s 
Since  s  increases  as  t  increases,  a  =  — -. 

dt 

d2s      * 

dv     m 
296 


RECTILINEAR  MOTION  297 

To  express  v  and  s  in  terms  of  t,  we  may  proceed  as  follows : 

d2s      f 
Since  —  =  ^-, 

ar     m 

.-.  —  =  *-  t  +  c,  by  integration. 
c?£     m 

To    determine   c :    When    £  =  0,   —  =  the   initial   velocity. 

Call  it  v0. 

.-.   i?0  =  0-f-c.         .*.  c  =  v0. 

at     m 

.-.    v=£*  +  Vo.  (2) 

m 

By  a  second  integration,  we  get 

s  =  %£-?  +  vQt  +  c1.  (3) 

m 

To  determine  cx :  When  t  =  0,  s  =  the  space  passed  over  before 
we  begin  to  count  time.  In  this  case,  therefore,  s  =  0  when 
(  =  0. 

m 

To  determine  v  in  terms  of  s,  we  may  express  £  in  terms  of  v 
from  (2),  substitute  the  result  in  (3),  and  solve  the  resulting 
equation  for  v.  It  is  more  convenient,  however,  to  proceed  as 
follows : 

Multiply  (1)  by  2  |. 

2  (tefdrs\  _ty  f  ds 
dt\dt2J~     mdt' 

Now  2^^=^-Y- 

dt\dt2J     dt\dtj 

d  fds\2_  r>  /  cZs 
dt\dt)         m  dt 


298       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

fcls\2  _o  / 


Integrate. 


...     ™)  =2J-s  +  c. 
\cltj         m 

m 


To  determine  c :  when  s  =  0,v=v0.     .'.  c  =  v0. 

m 


272.  As  an  illustration  of  motion  under  a  constant  force, 
consider  the  following  example : 

A  particle  of  mass  m,  initially  at  rest,  slides  down  a  smooth 
inclined  plane  of  length  a,  which  makes  an  angle  <£  with  the 
horizontal.     To  determine  the  velocity  of  the  particle  at  the 

foot  of  the  plane,  and  the  time  which 
the  particle  takes  to  reach  the  foot  of 
the  plane. 

Take  0  (Fig.  109),  the  initial  position 
of  the  particle,  as  the  origin,  and  the 
direction  of  motion  as  the  direction  of 
the  axis. 

The    only   force    tending    to    produce 
motion  is  the  component  weight  of  the 
particle  acting  along  the  plane.     This  force  is  mg  sin  <f>,  where 
g  depends  on  the  unit  system  chosen. 
The  equation  of  motion  is  therefore 


Fig.  109. 


or 


d2s 
m—=mgsm<l>, 

d2s  .     , 


(i) 


To  determine  the  velocity  at  the  foot  of  the  plane,  multiply 
(1)  by  2  — ,  integrate,  and  determine  the  constant  of  integration. 

KM) 

.-.  v2  —  2gs  sin  c/>.  (2) 


RECTILINEAR   MOTION  299 

Therefore,  at  the  foot  of  the  plane, 

v2  =  2  ga  sin  <f>,  or  v  =  V2  ga  sin  <£. 

To  determine  the  time  which  the  particle  takes  to  reach  the 
foot  of  the  plane,  we  may  proceed  as  in  the  last  article,  or  we 
may  integrate  (2)  directly.     Choosing  the  latter  way,  we  have 

ds 


Separate  the  variables. 


—  =  V2  as  sin  <f>, 
dt  *  * 

=  V2  g  sin  <£  •  Vs. 


=  V2  a  sin  d>  dt. 


Integrate,  and  determine  the  constant  of  integration. 
.*.  2  Vs  =  V2  g  sin  <£  t. 
.-.  s  —\g  sin  <f>  f. 

ATTRACTIVE  FORCE  VARYING   DIRECTLY  AS  THE  DISTANCE 

273.  Suppose  that  a  particle  of  mass  m,  initially  at  rest,  is 
attracted  towards  a  fixed  point  with  a  force  varying  directly  as 
the  distance.     To  determine  the  motion  of  the  particle. 

Let  0  (Fig.  110),  the  position  of  the  fixed  point,  be  the 
origin,  and  A,  distant  a  from  0,  the  initial  position  of  the  par- 
ticle.    Let  the  opposite  direction  to  the  motion  at  the  start 

be  the  positive  direction  of  the  axis. 

Let  s  be  the  abscissa  of  the  particle    °  0       s       P   ^ 

at  any  time  t.  FlG-  110' 

We  shall  suppose  as  before  that  m,  force,  and  a  are  expressed 
in  terms  of  the  units  of  mass,  force,  and  acceleration  resjjec- 
tively,  in  some  chosen  system  of  units. 

The  equation  of  motion  of  the  particle  between  A  and  0  is 

ma  =  Jcs,  (1) 

where  Jc  is  a  constant  at  present  undetermined. 


/ 


300       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

To  determine  k,  we  must  know  the  magnitude  of  the  attrac- 
tive force  or  acceleration  at  some  point  of  the  path.  Suppose 
for  definiteness  that  the  magnitude  of  the  acceleration  at  unit 
distance  from  0  is  known.  Suppose  that  it  is  /a.  Then 
m/A  =  ft  •  1.     .*.  k  =  m/x.     Equation  (1)  is  therefore 

ma  =  m/j.s, 

or  a  =  fxs.  (2) 

d2s 

Since  s  decreases  as  t  increases,  a  = ^r- 

dt2 

Equation  (2)  is  therefore  —  =  —[xs.  (3) 

LLC 

Equation  (3)  is  the  equation  of  motion  of  the  particle  from 
A  to  0. 

Multiply  (3)  by  2  — ,  integrate,  and  determine  the  constant 
of  integration.  „  y 

...   *  =  _VmV?=?,  (4) 

(1 9  • 

the  minus  sign  being  taken  because  —  is  negative,  since  s  de- 
creases as  t  increases. 

Separate  the  variables  in  (4). 


ds 


—  ^fjxdt. 


Va2  —  s2 
Integrate,  and  determine  the  constant  of  integration. 

•   -is  r~  *  i  v 

.\  sin  1-  =  -VwHr 
a  2 

.-.  s  —  a  sin  [  —  V/I  t  +  - 

.-.  s  =  a  cos  V/x  t.  (5) 

Differentiate  (5)  with  respect  to  t. 

.-.   —  =  —  a  V/x  sin  -\fjx  t.  (6) 

dt 


RECTILINEAR  MOTION 


301 


Equations  (4),  (5),  and  (G)  give  the  relations  between  each 

pair  of  the  variables  v,  s,  and  t. 

(Is  r~ 

From   equation    (4)  we   see  that  when  s  =  0,  -j-  =  —  a  V/x. 

Since  the  particle  has  therefore  a  velocity  at  0  and  there  is  no 
force  acting  on  it  there,  it  will  move  past  0.  At  any  distance 
s  to  the  left  of  0  there  is  a  force  the  same  in  magnitude  but 
opposite  in  direction  acting  on  it  as  acted  at  the  distance  s  to 
the  right.  We  see  therefore  that  the  motion  will  be  checked 
as  rapidly  as  it  was  increased,  and  that  the  particle  will  move 
to  a  point  B,  such  that  OB  =  —  a,  in  the  same  time  as  it  took  to 
reach  0  from  A.  At  B  the  same  force  is  acting  on  the  particle 
as  acted  before  at  A.  The  particle  will  therefore  return  to  A, 
moving  exactly  as  it  did  before,  and  continue  oscillating  be- 
tween A  and  B  in  equal  times. 

The 

2  7T 


From  equation  (5),  we  see  that  when  s  =  0,  t  = 


TV 


2V/x 


time  taken  for  the  particle  to  return  to  A  is  therefore  t  — 

This  result  is  remarkable  inasmuch  as  it  is  independent  of  the 
initial  distance  of  the  particle  from  0,  and  depends  only  on 
the  intensity  of  the  attraction  at  O. 

274.  Definition.  When  a  body  oscillates  between  two  points, 
the  time  of  a  complete  oscillation  is  called  the  periodic  time 
of  the  body. 

In  the  above  example  the  periodic  time  is  T=—-=- 

VjU, 

275.  On  BA  of  Fig.  110  as  diam- 
eter describe  a  circle  (see  Fig.  111). 
From  P,  the  position  of  the  parti- 
cle at  a  time  t,  draw  a  perpendic- 
ular PM  to  meet  the  circle  in  M. 
Denote  the  angle  POM  by  0. 

Therefore    s  =  a  cos  0. 

Also  s  =  a  cos  V/x  t. 

■      .'.   $  =  V/xt.  (1)  Fig.  111. 


302       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Equation  (1)  shows  that  if  a  particle  is  acted  upon  by  an 
attractive  force  varying  directly  as  the  distance  from  a  fixed 
point,  its  position  P  at  any  time  t  is  the  projection  on  the  line 
BA  of  the  point  M,  which  moves  with  the  uniform  angular 
velocity  V/x  on  the  circumference  of  the  circle  AMB. 


276.    Definition.     If  a  body  moves  in  a  straight  line  in  such 

dt2 
positive  constant,  or  can  be  reduced  to  this  form,  it  is  said  to 

have  simple  harmonic  motion. 


as 
a  way  that  its  equation  of  motion  is  —  =  —  /xs,  where  /a  is  a 


277.  It  has  been  determined  that  the  tension  in  a  stretched 
elastic  string  varies  directly  as  the  extension  of  the  string 
beyond  its  natural  length.  This  principle  is  known  as  Hooke's 
Law. 

From  Hooke's  Law  it  follows  that,  if  a  particle  moves  under 
the  force  exerted  on  it  by  a  stretched  elastic  string,  the  motion 
is  simple  harmonic  motion. 

ATTRACTIVE  FORCE   VARYING   INVERSELY  AS   THE   SQUARE 

OF  THE  DISTANCE 

278.  Suppose  that  a  particle  of  mass  m,  initially  at  rest,  is 
attracted  towards  a  fixed  point  with  a  force  varying  inversely 
as  the  square  of  the  distance.  To  determine  the  motion  of 
the  particle. 

Let  0  (Fig.  112),  the  position  of  the  fixed  point,  be  the 
origin,  and  A  distant  a  from  0,  the  initial  position  of  the  par- 
ticle.    Let  the  opposite  direction  to  the  motion  at  the  start  be 

the    positive    direction    of    the    axis. 

r 1 1 1  *■ 

B  0     s  A    ^gt  s  1^.  fc^  abscissa  of  the  particle 

Fig.  112. 

at  any  time  t. 

With  properly  chosen  units  for  mass,  force,  and  acceleration, 

the  equation  of  motion  of  the  particle  between  A  and  0  is 

m«=T  (1) 


UECTILINEAR  MOTION  303 

To  determine  k:  Suppose  that  the  magnitude  of  the  accelera- 
tion at  unit  distance  from  0  is  fi. 

ma 
Equation  (1)  is  therefore  m«  =  — g-> 

o 

or  ct  =  -o-  (2) 

Since  s  decreases  as  t  increases,  a  =  — —  • 

dv 

Equation  (2)  is  therefore  —-  =  —  £-.  (3) 

dtf         s2 

Equation  (3)  is  the  equation  of  motion  of  the  particle 
between  A  and  0. 

Multiply  (3)  by  2—,  integrate,  and  determine  the  constant 
of  integration. 

"*Y=2    (--- 


,.=— VS7lVg-5>  <*> 


ds 
dt 


the  minus  sign  being  taken  because  —  is  negative,  since  s  de- 
creases as  t  increases. 

Separate  the  yariables  in  (4),  integrate,  and  determine  the 
constant  of  integration. 


M  2  ii.\  *  a 


2  fx\_  *a 

As  the  particle  approaches  0  the  force  acting  on  it  increases 
without  limit.  Also,  as  seen  from  (4),  the  velocity  of  the  par- 
ticle increases  without  limit.  If  we  assume  that  when  the 
particle  is  at  0  the  force  acting  on  it  is  zero,  the  particle  will 
move  past  0.  At  any  distance  s  to  the  left  of  0  there  is  a 
force  the  same  in  magnitude  but  opposite  in  direction  acting 
on  it  as  acted  at  the  distance  s  to  the  right.  We  can  there- 
fore, on  this  assumption,  conclude,  as  in  Art.  273,  that  the  par- 


304       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

tide  will  move  to  a  point  B,  such  that  OB  =  —  a,  in  the  same 

time  as  it  took  to  reach  0  from  A,  and  that  it  will  oscillate 

between  B  and  A  in  equal  times.  r-— 

In  such  a  case  the  periodic  time  is  T=  ira\l 

*  /* 

CASE  OF  MOTION  IN  A  RESISTING  MEDIUM 

279.  Suppose  that  a  particle  of  mass  m,  initially  at  rest, 
moves  under  a  constant  force,  f,  in  the  direction  of  its  motion, 
and  in  a  resisting  medium  in  which  the  resistance  varies  as 
the  square  of  the  velocity.  To  determine  the  motion  of  the 
particle. 

Take  0  (Fig.  113),  the  initial  position  of  the  particle,  as  the 

origin,- and  the  direction  of  the  motion  as  the  positive  direction 

__  of  the  axis.     Let  s  be  the  abscissa  of 

' ' — — — p — ' — 

0  v  the  particle  at  any  time  t. 

Fig.  113.  r  J 


Since  s  increases  as  t  increases,  a  =  j 


dt2 


If  the  units  of  mass,  force,  and  acceleration  are   properly 
chosen, 

-£-'-<!)■  (1> 

To  determine  7c,  suppose  that  the  velocity  which  the  particle 
would  need  to  have  in  order  that  the  resistance  might  be  equal 

f 
to  /  is  ix.     Then,  /=  V-     •*■  fe  =  ~% ' 

Equation  (1)  is  therefore 


d2s       f 


dt2      m/x" 


ds 


(2) 


It   is   immediately  noticeable   that  equation  (2)  cannot   be 

integrated  by  multiplication  by  2—.     We  can  integrate,  how- 
ever, as  follows: 


RECTILINEAll  MOTION  305 


/7  a 

Let  v=  — ,  and  substitute  in  (2). 
eft 


.  dv_   f 


dt      wifx2 


IX2 -V2 


(3) 


Separate  the  variables  in  (3),  integrate,  and  determine  the 
constant  of  integration. 

1    ,        fX  +  v  f 

.'.  rr-  log™ =  ^—2t. 

Z  jx        fx  —  v      m/T 

e"  +  1 

where  I  =  —  • 

m/x 

From   equation   (4),  s  can  be   determined   immediately  in 

ds 

terms  of  t  by  substituting  —  for  v,  and  integrating. 

dt 


EXERCISES 

Assume  the  resistance  of  the  air  zero  unless  otherwise 
specified. 

1.  A  body  is  projected  vertically  downwards  from  the  top 
of  a  tower  200  feet  high  with  a  velocity  of  10  feet  per  second. 
Find  the  time  which  it  takes  to  reach  the  ground  and  its 
velocity  as  it  reaches  the  ground. 

Ans.  3.24  sees. ;  113.7  ft.  per  sec. 

2.  The  same  as  exercise  1,  excepting  that  the  body  is  pro- 
jected vertically  upwards  instead  of  vertically  downwards. 

Ans.  3.86  sees.;  113.7  ft.  per  sec. 

3.  A  balloon  is  ascending  with  a  velocity  of  20  miles  per 
hour.  A  man  in  the  balloon  drops  a  stone  which  reaches  the 
ground  in  6  seconds.  Find  the  height  of  the  balloon  when 
the  stone  is  dropped.  Ans.  400  ft. 


306       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

4.  A  particle  of  mass  m  is  projected  up  a  smooth  plane  in- 
clined at  an  angle  <f>  with  the  horizontal,  with  a  velocity  of  v0 
feet  per  second.  Find  how  high  it  will  ascend  the  plane  and 
the  time  it  takes  to  return  to  its  initial  position. 

Am.       v"       ft.;  -^-  sees. 
2  g  sin  cj>         g  sin  <j> 

5.  Show  that  the  times  of  descent  of  a  body  down  all 
chords  drawn  from  the  highest  point  of  a  vertical  circle  are 
equal  and  the  same  as  the  time  of  falling  down  the  vertical 
diameter. 

6.  A  train,  running  at  15  miles  per  hour  with  steam  shut 
off,  strikes  an  up  grade  of  1  in  300.  The  resistance  due  to  the 
air  and  friction  is  8  pounds  per  ton.  Find  how  far  the  train 
will  run  up  the  grade  before  coming  to  rest.      Am.  1031.25  ft. 

7.  In  the  preceding  exercise,  if  the  grade  were  a  down 
grade,  how  far  would  the  train  run  before  coming  to  rest? 

Am.  11343.8  ft. 

8.  On  the  moon  a  pound  weighs  2§  ounces.  On  the  earth 
a  man  can  jump  5  feet  high.  How  high  could  he  jump  on  the 
moon  ?  Ans.  30  ft. 

9.  A  cable-car  weighs  12  tons.  Find  the  tension  (assum- 
ing it  uniform)  in  the  cable  if  the  car  attains  a  velocity  from 
rest  of  12  feet  per  second  in  40  seconds. 

Ans.  A  force  of  225  lb. 

10.  A  train  rounds  a  curve  in  the  form  of  an  arc  of  a  circle 
of  500  feet  radius  at  the  rate  of  30  miles  per  hour.  Find  the 
angle  which  the  thread  of  a  plummet  suspended  from  the  roof 
of  the  car  makes  with  the  vertical.  Ans.  6°  54'. 

11.  A  car  starts  from  rest  with  an  acceleration  of  4  ft.-per- 
sec.  per  sec.  At  what  angle  must  a  man  brace  himself  to  keep 
in  equilibrium  ?  Ans.  7°  S'  (nearly). 


RECTILINEAR   MOTION  307 

12.  A  spiral  spring  of  natural  length  12  inches  is  stood  on 
one  end.  A  weight  of  15  pounds  placed  on  top  of  it  will  just 
hold  it  compressed  to  9  inches.  If  the  weight  be  placed  on  top 
of  it  when  in  its  natural  length  and  then  released,  determine  the 
motion. 

Suggestion.     The  spring  is  deformed  in  accordance  with  Hooke's  Law. 

Ans.  The  weight  descends  until  the  spring  is  compressed  to 
6  inches,  ascends  till  the  spring  is  of  natural  length, 
and  keeps  this  motion  up  indefinitely. 

o 

Incomplete  oscillation)  =  — -^-  =  .56  sec. 

13.  A  rubber  string  of  natural  length  2  feet  is  suspended  by 
one  end.  A  weight  of  ^  pound  attached  to  the  other  end  will 
just  hold  the  string  stretched  to  2\  feet.  If  after  the  weight 
is  attached  the  string  be  stretched  to  3  feet  and  then  released, 
determine  the  motion. 

Ans.  The  weight  ascends  till  the  string  is  of  natural  length, 
descends  till  the  string  is  stretched  to  3  ft.,  and 
keeps  this  motion  up  indefinitely. 

T=?-^  =  .78sec. 
8 

14.  A  round  spar  of  indefinite  length,  radius  of  cross  section 
a,  and  mass  M  floats  upright  in  water.  After  it  comes  to  rest 
it  is  pushed  down  a  distance  of  h  feet  and  then  released. 
Determine  the  motion. 

Ans.  It  rises  h  ft.  out  of  the  water  beyond  mark  when  at 
rest,  descends  h  ft.  below  mark  when  at  rest,  and 
keeps  this  motion  up  indefinitely. 

T=^-  where  fc  =  1000** 

15.  A  chain  64  feet  long  is  hung  over  a  smooth  drum,  one 
end  being  2  feet  lower  than  the  other.     Determine  the  velocity 


308       DIFFERENTIAL  AND  INTEGRAL    CALCULUS 

of  the  chain  when  one  end  is  12  feet  lower  than  the  other,  and 
the  time  taken  to  reach  this  velocity. 

Ans.   5.92  ft.  per  second ;  2.48  sees. 

16.  A  spring  which  offers  a  resistance  of  20  pounds  per  inch 
of  compression  stands  upright  on  a  fixed  plane.  A  weight  of 
10  pounds  falls  6  feet  and  strikes  the  spring.  How  far  will 
the  weight  descend  ?  Ans.   6|  ft. 

17.  A  ship,  whose  mass  is  1000  tons,  moves  from  rest  under 
a  constant  force  of  10  tons  and  is  resisted  by  the  water  with  a 
force  proportional  to  the  velocity  and  equal  to  2000  poundals 
when  the  velocity  is  1  foot  per  second.  Find  the  acceleration, 
and  the  distance  passed  over,  when  the  velocity  is  10  miles  per 
hour. 

Ans.  Accel.  =  0.305  ft.-per-sec.  per  sec. ;  $=343  ft.  (approx.). 

18.  A  weight  of  160  pounds  is  placed  on  a  plane  inclined 
20°  to  the  horizontal.  The  resistance  due  to  friction  is  5.72 
pounds.  The  resistance  due  to  the  wind  varies  as  the  square 
of  the  velocity,  and  is  2  pounds  per  square  foot  of  surface  when 
the  velocity  is  30  feet  per  second.  Given  that  the  area  exposed 
to  the  wind  is  4±-  square  feet,  find  the  velocity  when  the  weight 
has  descended  400  feet.  Find  the  limiting  value  of  the  veloc- 
ity on  a  hill  of  indefinite  length.  Ans.  62.54  ft. ;  70  ft. 


CHAPTER   XXXIII 


MOTION  m  A  PLANE   CURVE 


280.  The  simplest  case  of  motion  in  a  plane  curve  is  that  of 
a  particle  which  has  an  initial  velocity  and  is  acted  upon  by  a 
force,  constant  in  magnitude,  and  in  direction  constant  and 
oblique  to  that  of  the  initial  velocity.  Such  a  case  is  that  of  a 
projectile,  if  the  resistance  of  the  air  be  neglected,  because  for 
distances  through  which  a  projectile  carries,  the  force  of  gravity 
may  be  considered  as  constant  both  in  magnitude  and  direction. 

281.  Suppose  that  a  particle  of  mass  m  has  an  initial  velocity 
v0  which  makes  an  angle  a  with  the  horizontal,  and  is  acted 
upon  by  gravity  alone.  To  deter- 
mine the  motion  of  the  particle. 

Take  the  initial  position  of  the 
particle  as  the  origin  of  coordinates. 
Take  the  axes  vertical  and  hori- 
zontal.    (Fig.  114.) 

Since  there  are  no  forces  acting  on 
the  particle  in  the  direction  of  the 
z-axis,  d2x 


Fig.  114 


dt2 


=  0. 


(1) 


Since  the  force  of  gravity  acts  in  the  negative  direction  of  the 
?/-axis, 

(2) 


dhi_ 


-£  =  -9- 


Integrate  equation  (1). 


dx 
dt 


=  c. 


309 


310       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


To-  determine  c :  At  0  the  component  of  the  velocity  along 
the  x-axis  is  v0  cos  a. 


-.  c  =  v0  cos  a. 
dx 


dt 


=  v0  cos  a. 


(3) 


Integrate  equation  (3),  and  determine  the  constant  of  inte- 
gration. ,  , . . 


.-.  aj  =  v0cos  a  t 


Similarly,  from  equation  (2),  y  =  —  ±gt2  +  v0  sin  a  t.  (5) 

Substitute  the  value  of  t  from  (4)  in  (5). 


■••  y 


2  Vr?  cos2  a 


" — —  x2  +  x  tan  a. 


(6) 


Equation  (6)  is  the  equation  of  the  curve  which  the  particle 

describes  when  moving  under  the  above  supposition.     It  is  the 

,.         .  ,    ,        .,,    .,  ,         ,  /W  sin  2  a   v02sin2a 

equation  oi  a  parabola  with  its  vertex  at  ' 

j  £  i.  fv{)2  sin  2  a       v<?  cos  2  « 

and  focus  at    — 


2g    ' 


2g 


2g    '     2g   / 


MOTION  OF  A  PARTICLE  ON  A  SMOOTH  VERTICAL  CURVE 

282,   Let  y  =/(#)  be  the  equation  of  the  curve  on  which  the 
particle  moves.     Let  s  be  the  length  of  the  arc  between  some 

fixed  point  A  on  the  curve  and 
the  position  P  of  the  particle  at 
the  time  t.  Let  cf>  be  the  angle 
which  the  tangent  line  to  the 
curve  at  P  makes  with  the  a>axis. 
The  forces  acting  on  the  par- 
ticle are  the  force  of  gravity 
mg,  acting  vertically  downwards, 
and  the  resistance  R  of  the  curve 
FlG"  115'  acting  normal  to  the  curve. 

Since  the  resistance  acts  normal  to  the  curve,  it  does  not 
influence  motion  along  the  curve.     The  only  force  that  contrib- 


MOTION  IN  A  PLANE  CURVE  311 

utes  to  motion  along  the  curve  is  therefore  the  component  of 

the  force  of  gravity  along  the  curve.     The  equation  of  motion 

is  therefore 

d2s  •     , 

drs  •      ,  /in 

or  'dj?==~g        *'  '  ^ 

Now,  sin  <f>=  — ,  since  tan  </>  =  -£. 
ds  dx 


Multiply  by  2  — . 

(Jjv 


d2s  _    _     dy 
dt2  ds 


dt\dt)  U  dt' 


Integrate.  .-.  r-~j  =  -2gy  +  c. 


To  determine  c :  Suppose  that  the  body  is  at  rest  at  a  point 
A,  and  that  the  ordinate  of  A  is  h.     Then  c  =  2  gli. 

,(|J=2,(ft-,).  (2) 

To  integrate  further  we  must  know  y  in  terms  of  s,  and 
therefore  in  terms  of  x.  That  is,  we  must  know  the  equation 
of  the  curve. 


283.    Suppose  that  the  curve  of  the  preceding  article  is  the 
inverted  cycloid  whose  equations  are : 

x  =  a  (0  +  sin  6),  y  =  a(l  —  cos  6). 
Then  dx  =  a  (1  -+-  cos  6)  dd,  and  dy  =  a  sin  6  dO. 


,-.  ds  =  Va2 (1  +  cos  0)2  +  a2 sin2  6dd  =  2acosi0 dO. 


312        DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Therefore,  if  s  is  measured  from  the  origin, 

s  =  2  a   I     cos  \  6  d6  =  4  a  sin  -J-  6. 
Since         a  (1  —  cos  0)  =  2  a  sin2  J 6,     .'.  y  =  2  a  sin2  \B. 

8a 

Substitute  this  value  for  ?/  in  equation  (2)  of  the  preceding 

article"  .    /^Y=2     (h--S-\ 

"  \dt)         9\       SaJ 


=  -V2gyj} 


ds  h 


dt  *        8  a 

Separate  the  variables,  and  integrate. 

— =-^ 

V8a/i  V4ct 

To    determine   c  :    When   t  =  0,   y  =  h,   and   .-.   s  =  V8  a/i. 


-  =  _-*/ JL  t  +  C. 


7T 

~~2      .    c^-i      s 


9,         Via 


sm_1  — = \/-iZ- 1. 

.-.  s  =  V8a/i  sinf  ^—  . 

V2       '4  a 


=  V8  ah  cos 


Ha 


It  can  readily  be  shown  that  the  particle  will  rise  to  a 
point  B  at  a  height  7*.  on  the  other  side  of  the  lowest  point 
of  the  arch  and  oscillate  between  A  and  B  in  equal  times,  the 
periodic  time  being  _ 

\g 

Since  T  does  not  involve  h,  it  is  independent  of  the  position 
of  A.  Hence  the  periodic  time  is  the  same  for  all  arcs  of  the 
cycloid.     For  this  reason  the  curve  is  called  the  tautochrone. 


MOTION  IN  A   PLANE  CURVE  313 

THE   CYCLOIDAL   PENDULUM 

284.  As  seen  in  Art.  117,  the  evolute  of  a  cycloid  is  an  equal 
cycloid  having  its  vertices  at  the  cusps  of  the  given  cycloid. 
Hence  if  a  heavy  particle  be  sus- 
pended from  a  point  0  (Fig.  116) 
by  a  string  of  length  half  the  length 
of  an  arc  of  this  cycloid,  and  the 
string  in  its  oscillations  be  made 
to  wrap  itself  on  a  solid  piece  in 
the  shape  of  the  evolute  of  the 
cycloid,  the  particle  will  describe  the  given  cycloid. 

An  arrangement  of  this  sort  is  called  the  cycloidal  pendulum. 

If  s  and  </>  be  measured  as  in  Art.  282,  the  equation  of 
motion  for  the  cycloidal  pendulum  will  be  the  same  as  in  the 
case  of  a  particle  sliding  down  a  smooth  curve  under  the  action 
of  gravity.  All  the  results  of  Art.  283  will  therefore  hold  in 
the  case  of  the  cycloidal  pendulum.  We  can  thus  conclude 
that  the  periodic  time  of  a  cycloidal  pendulum  for  any  length 

of  arc  through  which  the  particle  swings  is  T=4:ir'y  — 

It  is  the  property  that  the  periodic  time  is  independent  of 
the  length  of  the  arc  through  which  the  particle  swings  that 
makes  the  cycloidal  pendulum  interesting. 


THE   SIMPLE   PENDULUM 

285.  Suppose  that  a  particle  of  mass  m,  suspended  from  a 
point  by  a  string  of  length  a,  starts  from  the  lowest  point  of 
its  path  with  the  velocity  which  it  would  acquire  if  it  fell 
freely  from  rest  in  a  vacuum  through  a  distance  h.  To  deter- 
mine the  motion  of  the  particle. 

Let  C  be  the  point  from  which  the  particle  is  suspended 
(Fig.  117).  Let  0,  the  lowest  point  of  the  arc  of  the  circle 
through  which  the  particle  swings,  be  the  origin  of  coordinates. 


314       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


Let  s  be  the  distance  from  the  origin  to  the  position  P  of 

the  particle  at  the  time  t. 
Let  the  coordinates  of  P 
be  (x,  y).  Let  <j>  be  the 
angle  which  the  tangent 
line  to  the  curve  at  P 
makes  with  the  #-axis. 

The  force  acting  on  the 
particle  at  a  time  t  is 
mg  sin  cf>.  Therefore,  since 
it  acts  opposite  to  the 
motion, 

ma  =  —  mg  sin  <£, 

a  =  —  g  sin  <f>. 

dh 

dt2 


Fig.  117. 


or 


Since  s  increases  as  t  increases,  a  =  -— 

,7y.2 


d2s 
dt 


2  =  -gsm<l>. 


Since 

Then,  as  in  Art.  282, 


.     ,      dy 

ds 


cPs 
dt2 


dy 
*  ds 


(1) 


ds 
dt 


=  -2gy  +  c. 


To  determine  c :    The   velocity  which   the   particle  would 
acquire  in  falling  freely  from  rest  in  a  vacuum  through  a 

,\  c  =  2gh. 


distance  h  is  ^/2gli. 


17=2,(7,-,). 


(2) 


The  equation  of  the  circle  is  x2  +  y2  —  2  ay  =  0. 


,-.  ds  —  -dy. 
x   y 


dt)      x2  [dt 


a2       fdy\2_ 


(3) 


MOTION   IN  A   PLANE  CURVE  315 


Substitute  the  result  of  (3)  in  (2). 
dy  __  V2g 


.:t  =  -Z-  C  dy      (4) 

V2^Jq  V(h-y)(2ay-y*) 

is  the  time  the  particle  takes  to  reach  a  point  whose  ordi- 
nate is  y. 

The  expression  for  t  in  (4)  is  transformed  into  one  or  other 

of  the  forms  >-    _  , 

ax 


or 


2  r*  da? 


/— -  r 


0*J«      /A        2Vi      2  a  , 

X"  1     1 JT 


^ 


by  the  transformations  given  in  Art.  224. 

As  stated  in  Art.  225,  either  of  these  integrals  is  an  Elliptic 
Integral  of  the  First  Class. 

If  h  <  2  a,  the  particle  comes  to  rest  when  y  =  h.  (See  equa- 
tion (2).)  Denote  this  point  on  the  circle  by  A.  Denote  the 
angle  OCA  by  «. 

Then  h  =  a(l  —  cos  a)  =  2  a  sin2  J  «. 

Then,  when  h<2a,  —  =  sin24-«.     Therefore  in  this   case 

the  value  k  in  the  elliptic  integral  is  the  sine  of  the  half  of  the 
half  angle  of  the  whole  swing  of  the  particle. 

EXERCISES 

1.    Show  that  the  parabola  of  equation  (6),  Art.  281,  has  its 

vertex  at  /   2    •    o  2   •   2    ■ 

vnz  sin  2  a    iv  snr  a 


29      '       2g     / 

and  its  focus  at      fo'  sin  2  ",   -^cos2"' 

2<7  2g 


316       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

2.  Find  the  horizontal  range,  and  the  time  of  flight  for  this 

range,  of  the  projectile  discussed  in  Art.  281. 

A        -r,              vn2  sin  2  a       m-          £  n-   t  ,      2i?ft2  sin  a 
Ans.  Range  =  -5 Time  of  flight  =  — 9 

3.  Show  that,  for  a  maximum  horizontal  range,  the  initial 
direction  of  the  path  of  projection  must  be  45°.  Determine 
the  position  of  the  focus  for  this  range. 

Ans.    The  focus  bisects  the  line  that  represents  the  hori- 
zontal range. 

4.  Find  the  range  and  time  of  flight  of  a  projectile  for  a 

plane  inclined  /?  to  the  horizontal. 

».  2     r 

sin  (2  a  —  /3)  —  sin  (3 


Ans.  Range  = 


g  cos1'  {3 

Time  of  flight  =     2  V°     sin  (a  -  /3). 
g  cos  /3 

5.  For  given  values  of  v0  and  /?,  find  the  value  of  a  which 
makes  the  range  of  Exercise  4  a  maximum. 

Ans.  a  =  i(90°  +  /3). 

6.  In  the  cycloidal  pendulum,  find  when  the  vertical  velocity 
is  greatest. 

Ans.  When  one-half  the  vertical  distance  has  been  traversed. 

7.  What  must  be  the  length  of  the  string  in  the  cycloidal 
pendulum  in  order  that  it  may  beat  seconds  ?      Ans.  38.91  in. 

8.  What  must  be  the  length  of  the  string  in  the  simple 
pendulum  in  order  that  it  may  beat  seconds  ? 

Ans.  38.91  in.  for  small  oscillations. 


9.    A  simple  pendulum  swings  through  an  angle  of  180°. 

7/ 


Find  the  time  of  oscillation. 

Ans.  o.ivb 


9 


CHAPTER   XXXIV 


WOEK  AND  ENEEGY 


286.  Work.  Work  is  said  to  be  done  on  a  body  by  a  force 
when  the  point  of  application  of  the  force  has  a  component 
displacement  in  the  direction  of  the  force.  Work  is  said  to  be 
done  by  a  body  against  a  force  when  the  point  of  application 
of  the  force  has  a  component  displacement  in  a  direction  op- 
posite to  that  of  the  force.  In  the  latter  case  work  may  be 
said  to  be  done  on  the  body  by  the  force  if  distinction  be  made 
between  positive  and  negative  work. 

287.  Measurement  of  the  work  done  on  a  particle  in  a  given  dis- 
placement when  the  force  is  constant  in  magnitude  and  direction, 
and  motion  is  in  a  straight  line.  If  the  force  is  constant  in 
magnitude  and  direction,  and  motion  takes  place  in  a  straight 
line,  the  work  done  by  the  force  in  a  given  displacement  of 
the  particle  is  measured  by  the  product  of  the  force  and  the 
component  of  the  displacement  along  the  line  of  force,  the 
work  being  taken  positive  if  the  component  of  the  displace- 
ment is  in  the  direction  of  the  force,  and  negative  if  it  is  in 
the  direction  opposite  to  that  of  the  force. 


Fig.  118. 


s 
Fig.  119. 


Thus,  if  w,  F,  s,  and  a    (Figs.  118  and  119)  represent  the 
work  done,  the  force,  the  displacement,  and  the  angle  between 

317 


318       DIFFERENTIAL  AND  INTEGRAL    CALCULUS 

the  displacement  and  the  direction  of  the  force  respectively, 
*nen  io  x  Fs  cos  a. 

Since  iv  oc  Fs  cos  a,  w  =  k  Fs  cos  a,  where  k  is  a  constant  de- 
pending on  the  units  chosen  for  w,  F,  and  s.  As  in  Arts.  264 
and  265,  we  shall  find  it  convenient  to  choose  7c  equal  to  1. 
The  unit  of  work  will  therefore  be  that  work  done  when  the 
body  acted  upon  by  the  unit  of  force  has  a  displacement  of 
one  unit  in  the  direction  of  the  force. 

Since  F-  s  cos  a  =  Fcos  a  •  s,  the  work  done  is  also  measured 
by  the  product  of  the  component  of  the  force  in  the  direction 
of  the  displacement  and  the  displacement. 

288.  Units  of  Work.  The  systems  of  units  of  work  in  com- 
mon use  are  the  F.P.S.  and  C.G.S.  gravitational  systems,  and 
the  F.P.S.  and  C.G.S.  absolute  systems. 

In  the  F.P.S.  gravitational  system,  the  unit  of  work  is  the 
work  done  when  a  force  of  one  pound  moves  a  body  through 
a  distance  of  one  foot  in  its  direction.  This  unit  of  work  is 
called  the  foot-pound. 

In  the  C.G.S.  gravitational  system,  the  unit  of  work  is  the 
work  done  when  a  force  of  one  gram  moves  a  body  through 
a  distance  of  one  centimeter  in  its  direction.  This  unit  of 
work  is  called  the  gram- centimeter. 

In  the  F.P.S.  absolute  system,  the  unit  of  work  is  the  work 
done  when  a  force  of  one  poundal  moves  a  body  through  a 
distance  of  one  foot  in  its  direction.  This  unit  of  work  is 
called  the  foot-poundal. 

In  the  C.G.S.  absolute  system,  the  unit  of  work  is  the  work 
done  when  a  force  of  one  dyne  moves  a  body  through  a  dis- 
tance of  one  centimeter  in  its  direction.  This  unit  of  work 
is  called  the  erg. 

289.  Measurement  of  the  work  done  on  a  particle  in  a  given 
displacement  when  the  force  is  variable  in  magnitude  or  direction 
or  both. 


WORK  AND  ENERGY 


319 


First,  when  motion  is  in  a  straight  line. 

Let  AB  (Fig.  120),  where  A  and  B  have  the  abscissas  a  and 
b  respectively,  be  the  given  displacement. 

Divide  AB  into  n  equal  parts.     Call  each  part  As. 

Let  F'k  and  F"k  denote  the  least  and  greatest  values  respec- 
tively of  the  force  that  acts  on  the 
particle  in  an  interval  As  of  the  dis- 
placement. Let  <f>'k  and  <j>"k  denote 
the  greatest  and  least  values  respec- 
tively of  the  angle  which  the  force 
makes  with  the  displacement  in  this  interval.  Let  wk  denote 
the  work  done  by  the  force  in  the  interval. 


0 


Fig.  120. 


Then 


F\  cos  </>'„  As  <  wk  <  F"k  cos  <j>"k  As. 


(1) 


Let  Fk  denote  the  force  at  the  beginning  of  the  interval  As, 
and  cf>k  the  angle  which  the  force  makes  with  the  displacement 
at  this  point.     Divide  (1)  by  Fk  cos  <f>k  As. 


.    *Vos*'fc 


< 


Wi. 


< 


F"kcos<J>"k 


Fkcoscf>k      Fkcos<t>kAs       Fkcos<j>k 


Now 


limit 
As  =  0 


.     limit 
'  '   As  =  0 


limit 
As=0 


=  1. 


~F"k  cos  $" 
_  Fk  cos  <f,k  _ 


=  1. 


Therefore,  by  the  theorem  of  Art.  186,  wk  may  be  replaced  by 
Fk  cos  <f>k  As  in  any  problem  involving  the  limit  of  the  sum  of 
the  infinitesimals  wk. 


x=b 


.-.  w=limit  y  F  cos  6  As 


=  r\ 

%j  x=a 


Fcos  cfads. 


Second,  when  motion  is  along  a  curve. 

It  can  be  shown  by  a  method  which  does  not  differ  in  im- 


320       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

portant  details  from  that  given  above,  but  is  more  complicated 
that  the  work  done  in  this  case  is 


%J  X: 


x=b 

F  cos  <£  ds, 

x=a 


where  F  is  the  force  at  any  point  of  the  curve,  <£  the  angle 
which  the  force  makes  with  the  tangent  line  to  the  curve  at 
this  point,  and  a  and  b  the  abscissas  of  the  initial  and  final 
points  respectively  of  the  displacement. 

We  shall  assume  the  theorem  without  discussion. 

290.  Work  done  by  a  resultant.  If  any  number  of  forces  act 
on  a  particle,  the  algebraic  sum  of  the  work  done  by  the  forces 
acting  separately  is  the  same  as  if  the  resultant  of  these  forces 
alone  acted  on  the  particle. 

Let  B  be  the  resultant  of  n  forces  flf  /2,  j^,  •••,  fn,  which  act 
on  a  particle.  Since  B  is  the  closing  line  of  the  polygon  formed 
by  adding  the  forces  geometrically,  its  projection  on  any  line 
is  the  sum  of  the  projections  of  the  forces  f1}  f2,  f3,  •••,  fn.  That 
is,  if  B,fi,f2,fs,  -~,fn,  make  angles  of  0,  01}  62,  03,  ••-,  0n  respec- 
tively with  a  line, 

B  cos  0  =f1  cos  0!  +/a  cos  02  +f8  cos  03  H \-fn  cos  0n. 

Let  the  line  with  which  these  forces  make  these  angles  be 
the  tangent  line  to  the  path  of  displacement  of  the  particle  at 
any  point.  Therefore,  by  Art.  289,  the  work  done  by  B  between 
two  points  A  and  B  on  the  path  whose  abscissas  are  a  and  b 
respectively  is 

J-*x=b  S*x=b 

B  cos  0  ds,  or   I       (/x  cos  0j  +/2  cos  02 
x=a  %J  x=a 

+/3  cos  03  H h/n  cos  On)ds. 

But   \       (/x  cos  0!  +/2  cos  02  +/3  cos  03  H \-fn  cos  6n)  ds 

%J  x=a 

J-*x=b                                   r»x=b                                   r*x=b 
fi  cos  0!  ds  +  |       /a  cos  02  ds  +  I       /3  cos  03  ds  -\ 
x=a                                      *J  x=a                                       J  x=a 

J~*x=b 
fn  cos  0n  ds. 
x—a 


WORK  AND   ENERGY 


321 


Therefore  the  algebraic  sum  of  the  work  done  by  all  the  forces 
acting  separately  is  the  same  as  the  work  clone  if  the  resultant 
alone  acted  on  the  particle. 

291.  Central  Force.  A  force  which  is  always  directed  towards 
a  fixed  point  and  varies  as  the  distance  of  its  point  of  applica- 
tion from  the  fixed  point  varies  is  called  a  central  force. 

To  prove  that  the  work  done  by  a  central  force  is  independ- 
ent of  the  path. 

Since  a  central  force  varies  as  the  distance  of  its  point  of 
application  from  the  fixed  point  varies,  it  is  a  function  of  its 
distance  from  the  fixed  point.     Call  it/(r). 

Take  the  fixed  point  as  the  pole  and  express  the  equation  of 
the  path  in  polar  coordinates. 

Then 

tan^  =  r— •    (See  Art.  102.) 

Since 
cos  \p  = 


sec^      Vl+tan2^ 


.-.  cos  \\l 


4 


-*(f 


VcZr2  +  r2dd2 


.'.    COS  \f/  = 


clr 


(See  Art.  199.) 


ds 

.-.  dr  =  cos  ij/  ds. 

Therefore  the  work  done 

—  I  f(r)  c°s  (w  —  \j/)  ds  =  —  jf(r)  cos  ij/ds, 
=  -  ff(T)dr,  (1) 


the  limits  of  integration  in  the  last  integral  being  the  initial 
and  final  values  of  r. 

Since  (1)  is  a  function  of  the  initial  and  final  values  of  r  only, 
the  work  done  is  independent  of  the  path. 


322       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

292.  Work  expressed  in  terms  of  the  components  of  the  forces 
in  the  directions  of  the  axes.  Suppose  that  any  number  of  forces 
act  on  a  particle.  Let  X,  Y,  Z  be  the  sum  of  the  components 
of  these  forces  in  the  directions  of  the  axes.  Let  R  be  the 
resultant  of  the  forces,  and  0  the  angle  which  the  tangent  line 
to  the  path  of  displacement  of  the  particle  makes  with  R  at 
any  time.  Let  the  direction  cosines  of  the  tangent  line  be 
cos  a,  cos  {3,  cos  y.  Then  the  work  done  by  the  given  forces 
during  a  given  displacement  is 

J  R  cos  0  ds,  or  J  (X  cos  a  +  Y  cos  /?  +  Z  cos  y)ds. 

Now         cos  ads  =  dx,  cos  f3  ds  =  dy,  cos  yds  =  dz. 

.\  C(X  cos  a  +  Fcos  /?  +  Z cos  y)  ds=  C(Xdx  +  Ydy  -f  Zdz). 

Therefore  the  work  done  by  the  forces  during  a  given  displace- 
ment of  the  particle  is  |  (Xdx  -\-  Ydy  +  Zdz)  between  proper 
limits. 

293.  Rate  of  Work.  The  mean  or  average  rate  at  which  work 
is  done  by  a  body  is  the  work  done  in  a  period  of  time  divided 
by  the  time. 

The  rate  of  work  done  by  a  body  at  any  particular  instant 
is  the  limit  which  the  mean  or  average  rate  of  work  for  a  period 
of  time  immediately  succeeding  the  instant  in  question  ap- 
proaches as  the  period  of  time  is  allowed  to  become  indefinitely 
decreased. 

294.  Units  of  Rate  of  Work.  The  systems  of  units  of  rate  of 
work  derived  from  the  units  of  work  are  the  foot-pound  per 
second,  the  centimeter-gram  per  second,  the  foot-poundal  per 
second,  and  the  erg  per  second. 

The  unit  used  in  practice  by  American  engineers  is  the  horse- 
power, which  is  550  foot-pounds  per  second.     The  French  use 


WORK  AND  ENERGY 


323 


the  unit  75  kilogrammeters  per  second,  equivalent  to  542486 
foot-pounds  per  second,  which  they  call  the  force  de  cheval. 

A  unit  extensively  used  in  electrical  work  is  the  watt,  which 
is  10,000,000  ergs  per  second. 

295.  Energy.  As  stated  in  Art.  286,  a  body  is  said  to  do 
work  against  a  force  when  the  point  of  application  of  the  force 
has  a  component  displacement  in  a  direction  opposite  to  that 
of  the  force. 

Definition.  A  body  able  to  do  work  against  a  force  is  said 
to  have  work-power  or  energy. 

296.  Units.  Energy,  being  power  of  doing  work,  is  measured 
in  terms  of  the  units  of  work. 

297.  Kinetic  Energy.  If  a  body  has  a  velocity,  it  is  capable 
of  doing  work  against  a  force  that  has  a  component  opposite  to 
the  direction  of  the  velocity. 

Definition.  The  work-power  or  energy  of  a  body  due  to  a 
velocity  is  called  the  kinetic  energy  of  the  body. 

298.  Determination  of  the  kinetic  energy  expended  in  a  given 
displacement  of  a  particle.  Suppose  that  a  particle  of  mass  m 
has  an  initial  velocity  v0,  and  is  acted  upon  by  a  force  F,  which 
makes  an  angle  6  with  the  opposite  direction  of  the  line  of 
motion  of  the  particle.  To  determine  the  kinetic  energy 
expended  in  a  given  displacement  of  the  particle. 

Let  the  displacement  of  the  parti- 
cle be  from  A  to  B,  where  A  and  B 
have  the  abscissas  a  and  b  respec- 
tively (see  Fig.  122). 

The  equation  of  motion  of  the  par- 
ticle is  p 

m- —  =  —  FcosO, 
df 


or 


dn 
mv—  =  —  Fcos  0,  since  ^-  =  v  — 
ds  dt'~        ds 


324       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Integrate.  .-.  ^  =  -  JV  cos  0  ds  +  c. 

To  determine  c :  Suppose  for  a  moment  that 
JF7cos0ds  =  <£(x). 


/> 


mv 


,2 


When  cc  =  a,  v  —  v0,  and  <£(a?)  =  <f>(a).     .:  c  =  —^-  +  <f>(a). 


2 


>^2_^=_^)  +  ^(a). 


When  a  =  6,  v  =  the  velocity  at  B,  and  <f>(x)  =  <f>(b).     Denote 
the  velocity  at  B  by  vb. 

Jr*x-i 
F  cos  6  ds. 
x=a 

...  El_^=  f°Vcos0<fe. 


Now  I       -F7  cos  0  c?s  is  the  work  done  by  the  particle  against 

«y  x=a 

F  in  its  motion  from  A  to  5,  and  by  definition  this  is  the 
kinetic  energy  expended.  2  g 

Therefore  the  kinetic  energy  expended  is  — -  —  — -*-• 

When  the  particle  comes  to  rest,  i>6  =  0.  Therefore  the 
kinetic  energy  of  a  particle  =  \  mv2,  where  v  is  the  velocity 
of  the  particle. 

299.  Potential  Energy.  A  body  acted  upon  by  a  force  di- 
rected towards  a  center,  and  in  a  position  from  which  it  can 
move  towards  the  center,  can,  by  virtue  of  its  position,  do  work 
against  another  force  having  a  component  in  a  direction  oppo- 
site to  that  of  the  first  force.  Thus,  a  weight  in  a  position 
from  which  it  can  fall  towards  the  center  of  the  earth  can  do 
work  against  a  force  restraining  it. 


WORK  AND  ENERGY  325 

Definition.  The  work-power  which  a  body  possesses,  due  to 
its  position,  is  called  the  energy  of  position,  or  the  potential 
energy  of  the  body. 

300.  A  particle  will  possess  potential  energy  in  whatever 
position  it  may  be  placed  in  the  field  of  force,  but  the  nearer 
it  comes  to  the  center  of  force,  the  less  potential  energy  it 
possesses.  Suppose  that  a  particle  is  acted  upon  by  a  force  F, 
and  is  held  from  moving  towards  the  center  of  force  by  a  re- 
straining force  which  has  a  component  equal  to  F  and  in  the 
opposite  direction  to  it.  If  the  attractive  force  F  be  increased 
by  ever  so  small  an  amount,  the  work  done  in  moving  the  par- 
ticle from  a  point  A  to  a  point  B  at  a  distance  s  from  A  and 
in  the  line  of  force,  differs  by  ever  so  small  an  amount  from 

I  Fds,  between  proper  limits.     Then  I  Fds,  between  proper 

limits,  may  be  taken  as  the  work  done  in  moving  the  particle 
from  A  to  B.  Then  the  particle  loses  this  amount  of  potential 
energy. 

Similarly,  if  motion  is  in  the  opposite  direction  to  that  of 
the  attractive  force,  the  particle  gains  in  potential  energy  by 
the  amount  of  work  done. 

For  example,  it  follows  immediately  from  Art.  286  or  287 
that  the  amount  of  work  done  in  raising  a  weight  of  mass  m 
to  a  height  h  is  mgh.  Since  this  amount  of  work  is  mgh,  the 
potential  energy  gained  by  the  weight  by  being  raised  a  height 
h  is  therefore  mgh.  Similarly,  if  the  weight  be  lowered  through 
a  distance  h,  the  potential  energy  lost  is  mgh. 

EXERCISES 

1.  Convert  1  foot-pound  to  gram-centimeters. 

Ans.   1  foot-pound  =  13,824.7  gram-centimeters. 

2.  Convert  1  foot-poundal  to  ergs. 

Ans.    1  foot-poundal  421,358  ergs. 


326       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

3.  Convert  1  erg  to  grain-centimeters. 

Ans.   1  erg  =  1019  x  10-5  gram-centimeters. 

4.  A  train  weighs  100  tons.  The  resistance  due  to  the  air 
and  friction  is  8  pounds  per  ton.  Find  the  work  done  in  moving 
the  train  one  mile  up  a  grade  of  1  in  400. 

Ans.   6864000  foot-pounds. 

5.  A  cylindrical  cistern  30  feet  deep  and  6  feet  in  diameter 
is  full  of  water.  Find  the  work  done  in  emptying  the  cistern. 
(The  weight  of  a  cubic  foot  of  water  =  62i  lb.) 

Ans.   795200  foot-pounds  (approx.). 

6.  Show  that  when  a  piston  is  driven  by  the  pressure  of  an 
expanding  fluid,  the  work  done  when  the  volume  changes  from 

pdv,  where  p  is  the  pressure  per  unit  area  on  the 

piston.  Interpret  this  work  as  the  area  inclosed  by  the  curve 
p  =  f(v)}  the  v-axis,  and  the  ordinates  corresponding  to  the 
abscissas  v0  and  v1  respectively. 

7.  When  the  volume  of  steam  in  the  cylinder  of  the  steam 
engine  is  1.2  cubic  feet,  and  the  pressure  is  60  pounds  per 
square  inch  on  the  piston,  the  steam  is  cut  off.  Find  the  work 
done  when  the  volume  has  changed  to  3.2  cubic  feet ;  given 
that  the  volume  and  pressure  when  the  steam  is  cut  off  are  con- 
nected  by  the  equation^16  =  a  constant. 

Ans.   9854  foot-pounds. 

8.  Steam  is  admitted  into  the  cylinder  of  a  steam  engine  under 
a  constant  pressure  of  80  pounds  per  square  inch.  When  the 
volume  of  steam  in  the  cylinder  increases  from  0  to  2  cubic 
feet,  the  steam  is  cut  off.  The  steam  in  the  cylinder  then  ex- 
pands  under  the  law  pv  T^=  a  constant  until  the  volume  is  6 
cubic  feet.  It  is  then  expelled  under  a  constant  back  pressure 
of  10  pounds  per  square  inch.  Find  the  work  done  by  the 
steam  on  the  piston.  Ans.   38860  foot-pounds  (approx.). 


WORE  AND  ENERGY  327 

9.    Find  the  work  done  by  an  expanding  fluid  which  obeys 

4 

the  law  pv*  =  11520  when  v  changes  from  0.5  cubic  foot  to  2 
cubic  feet,  (j)  =  number  of  pounds  pressure  per  square  foot; 
v  =  number  of  cubic  feet.)  Ans.   16112  foot-pounds. 

10.  Defining  the  average  or  mean  value  of  f(x)  in  the  interval 
a  to  b  to  be  the  limit  when  n  =  oo  of  the  average  of  n  values  of 
f(x)  corresponding  to  equidistant  values  of  x  in  the  interval  a 
to  b,  show  that  the  mean  value  of  f(x)  in  the  interval  a  to  b  is 

j 
f(x)dx 

b  —  a 

11.  Find  the  mean  pressure  in  Exercise  7. 

Ans.   34.2  lb.  per  sqnare  inch. 

12.  Find  the  mean  effective  pressure  of  the  steam  in  Exer- 
cise 8.  Ans.   45  lb.  per  sqnare  inch. 

13.  A  variable  force  has  acted  through  2  feet.  The  value  of 
the  force  taken  at  seven  equidistant  points  including  the  first 
and  last  is,  in  pounds,  200,  176,  142,  108,  76,  54,  20.  Find,  by 
Simpson's  Rule,  an  approximation  to  the  work  done. 

Ans.    223.1  foot-pounds. 

14.  Investigate  Exercise  12,  Chapter  XXXII,  for  the  work 
done  nntil  the  spring  is  compressed  to  6  inches. 

Ans.   1\  foot-pounds. 

15.  Investigate  Exercise  13,  Chapter  XXXII,  for  the  work 
done  until  the  weight  begins  to  descend.       Ans.   \  foot-pound. 

16.  Convert  1  horse-power  to  watts. 

Ans.   1  horse-power  =  742  watts. 

17.  Convert  1  force  de  cheval  to  ergs  per  second. 

Ans.   1  force  de  cheval  =  7.36  x  109  ergs  per  sec. 

18.  A  mass  of  20  pounds  is  being  dragged  along  a  horizontal 
plane,  the  rate  of  work  done  being  -^  horse-power.  Find  its 
acceleration  when  its  speed  is  2  miles  per  hour. 

Ans.   30  ft.-per-sec.  per  sec.  in  the  direction  of  its  motion. 


328       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

19.  If  the  expansion  in  Exercise  9  above  takes  place  in  0.1 
second  what  is  the  horse-power  ?  Ans.   292.9  horse-power. 

20.  A  cannon  ball  weighing  200  pounds  is  discharged  with  a 
velocity  of  1200  feet  per  second.     Eind  its  kinetic  energy. 

Ans.    45  x  105  foot-pounds. 

21.  In  Exercise  13,  Chapter  XXXII,  find  the  kinetic  energy 
expended  while  the  string  contracts  from  3  to  2  feet.     Ans.  0. 

22.  A  stone  weighing  1  ton  is  lifted  30  feet.     Find  its  in- 
crease in  potential  energy.  Ans.   6  x  104  foot-pounds. 


CHAPTER   XXXV 

ATTKAOTION 

301.  Definitions.  The  mean  or  average  density  of  a  body  is 
the  mass  of  the  body  divided  by  the  volume. 

The  density  of  a  body  at  a  point  is  the  limit  which  the  mean 
or  average  density  of  an  element  of  volume  containing  the 
point  approaches  as  the  element  of  volume  is  allowed  to  be- 
come indefinitely  decreased. 

A  body  is  said  to  be  homogeneous  when  its  density  is  the 
same  at  every  point. 

A  body  not  homogeneous  is  said  to  be  heterogeneous. 

302.  Units  of  density.  The  unit  of  density  is  unit  mass 
divided  by  unit  volume. 

In  the  F.P.S.  system  there  is  probably  no  familiar  sub- 
stance whose  density  is  the  unit  density.  The  mass  of  a  cubic 
foot  of  water  is  approximately  62^-  times  that  of  the  ideal  sub- 
stance of  unit  density.  Then  in  the  F.P.S.  system,  the  density 
of  water  is  62|-  times  the  unit  density  in  this  system. 

The  gram  was  chosen  to  be  the  mass  of  a  cubic  centimeter 
of  distilled  water  at  4°  C.  Then  in  the  C.G-.S.  system  the 
unit  of  density  is  the  density  of  distilled  water  at  4°  C. 

303.  Every  body  in  the  universe  attracts  every  other  body 
with  a  force  which  depends  in  magnitude  and  direction  on  the 
masses  and  relative  positions  of  the  bodies. 

In  the  case  of  two  bodies  considered  as  particles,  the  force 
of  attraction  varies  directly  as  the  product  of  the  masses  and 
inversely  as  the  square  of  the  distance  between  the  particles. 
Thus,  if  F  is  the  force  of  attraction  between  two  particles  of 

329 


330       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

masses  m  and  m'  respectively,  and  r  is  the  distance  between 

the  particles,  mm' 

F<x — 5-- 

In  the  case  of  two  continuous  bodies,  the  force  of  attraction 
may  be  measured  approximately  by  dividing  the  masses  up 
into  infinitesimal  elements  of  mass,  forming  the  expression 

')lt  Til 

-~2~  for  each  pair  of  the  elements,  and  taking  the  sum  of  the 

results.  The  actual  force  of  attraction  between  the  two  bodies 
is  the  limit  which  this  sum  approaches  as  the  number  of  ele- 
ments is  allowed  to  increase  without  limit  while  each  element 
decreases  without  limit. 

304.  Newton's  Law  of  Gravitation.  The  above  law  of  gravi- 
tation, first  enunciated  by  Newton  in  the  case  of  particles,  was 
derived  by  him  from  certain  laws  —  called  Kepler's  laws  — 
observed  from  the  motions  of  planets,  and  may  be  accepted 
with  as  much  confidence  as  was  the  fundamental  law  of 
motion,  F=  ma. 

305.  Since  Foc—j-,F= — ^— ,  where  7c  is  a  constant  de- 
pending on  the  units  chosen  for  F,  m,  m',  and  r.  In  any  system 
of  units,  Jc  is  evidently  the  force  with  which  two  particles  each 
of  unit  mass  attract  each  other  when  at  unit  distance  apart. 

Since  F—~k—-^-  whatever  be  the  masses  of  the  attracting 
particles,  the  force  of  attraction  between  two  particles,  one  of 
mass  m,  and  the  other  of  unit  mass,  is  —j-  • 

Definition.  The  force  of  attraction  between  two  particles, 
one  of  mass  m  and  the  other  of  unit  mass,  is  called  the  force  of 
attraction  of  the  particle  of  mass  m  at  the  point  where  the 
unit  particle  is  situated. 

306.  Determination  of  k:  It  will  be  shown  later  (see  Art. 
311)  that  the  attraction  of  a  homogeneous  sphere  on  an  ex- 
ternal point  is  the  same  as  if  the  mass  of  the  sphere  were  con- 


ATTRACTION  331 


centrated  at  the  center.  Then,  assuming  this  theorem,  we 
have  that,  if  m  be  the  mass  of  the  earth,  assumed  a  homoge- 
neous sphere,  and  m'  the  mass  of  the  particle,  the  equation 

F—k^-j-  expresses  the  force  of  attraction  between  the  earth 

F         TCTYl 

and  particle.     Divide  the  equation  by  m\     .*.  — -f  =  —j-.     Now 

F  ,m       r      . 

— j  is  the  acceleration  of  the  particle  due  to  the  attraction  of 

k  a  2 

the  earth.     Call  it  q,  as  in  Art.  43.     .'.  a  —  — «-•     .\  k  =  —  • 

The  radius,  r,  and  the  density,  p,  of  the  earth  are  known  to  be 
6.37  x  108  cm.  and  5|  respectively.  In  the  C.G.S.  absolute 
system  of  units,  therefore, 

k  =  - — =  645  x  10-10. 

4  7T  x  3.281  x  5}  X  6.37  x  108 

In  this  system  of  units,  k  is  the  force  in  dynes  with  which 
two  particles  each  of  mass  one  gram  would  attract  each  other 
when  situated  at  a  distance  of  one  centimeter  apart. 

307.  Astronomical  System  of.  Units.  When  the  units  are  so 
selected  that  k  =  1,  the  unit  of  mass  is  that  mass  which  pro- 
duces the  unit  of  force  in  an  equal  mass  when  at  unit  distance 
from  it.  This  system  of  units  is  called  the  astronomical  sys- 
tem of  units. 

ATTRACTION  AT  A  POINT  DUE  TO  A  SYSTEM  OF  PARTICLES 

308.  Let  m„  m2,  m3,  •••,  mn  be  a  system  of  n  particles.  To 
find  the  attraction  at  a  point  0  due  to  these  particles. 

Let  the  origin  of  coordinates 
be   taken  at   0.     Let  aif  /?,-,  yt- 
be  the  direction  angles  of  the        %m 
particles  of   mass  raf,  %  =  1,  2, 

3,  ••-,  n.     Let  X,  Y,  Z  be  the    ' 

components     of    the    resultant  - 

B    in '  the    directions    of    the 

axes.  v/  Fia.  123. 


'm*      »m, 


m, 


332       DIFFERENTIAL  AND  INTEGRAL    CALCULUS 


,  mx  cos  ax      m9  cos  a2      wi3  cos  «3  mn  cos  «, 

Inen  A  = , 1 « ' ^ r  •••  H 5 — ! 

?\2  r/  r3J  r2 


1=  n 

1=1 

_m1cosff1     m2cos/?2     m3cosft3  mwcos/?n 

n2  r22  r32  rn2 


=x 


mt  cos  j3i 


i  =  \  • 


n2  r?  n2 


mjcosy!      m2cosy2     m3cosy3  mncosyn 

^  =  — r^ 1 n> — -  -\ Z72 r  '"  -\ zrr~ 

'1 

~y^\  mt  COS  yt 
=  2,  T2 


t=l 


The  resultant  attraction  is  R  =  VX2  +  F2  +  Z2,  and   the 

direction  cosines  of  its  line  of  application  are  — ,  — ,  and  — 

M     M  M 

respectively. 

ATTRACTION   AT  A  POINT   DUE  TO  A  STRAIGHT 
HOMOGENEOUS   WIRE 

309.    Let  0  be  the  given  point,  and  AB  the  given  wire. 
Choose  the  point  0  as  the  origin  of  coordinates,  and  the  per- 
pendicular from    0  on  AB  as   the 
direction  of  the  aj-axis. 

Let  c  denote   the  distance  of   0 
Afc+i  from  AB,  and  b  the  length  of  AB. 

Afc  Let  X  and  Y  be  the  components  of 

the  attraction  of  AB  in  the  direc- 
tions of  the  axes.  Denote  the 
density  of  the  wire  by  p. 


0 


^  We  shall  suppose  here  that  the 

perpendicular  from  0  on  AB  meets 
AB  in  A.  The  cases  where  it  does  not  are  left  as  exercises 
to  the  student. 


ATTRACTION  333 

Divide  AB  into  n  equal  parts.  Call  each  part  Ay.  Let  Ak 
and  Ak+1,  with  ordinates  yk  and  yk  +  Ay  respectively,  be  two 
successive  points  of  division  of  AB. 

The  resultant  attraction  at  0,  due  to  the  element  AkAk+1, 

•  -,      -,  i  A'pA?/  .,  .        J .         &pA?/ 

is  evidently  greater  than  -s — ,      '~       x9  and  less  than  -=— — o* 
J  &  c2  +  (y*  +  Ay)2  c2  +  2/t2 

The  direction  of  the  resultant  attraction  due  to  AkAk+l   is 

along   a   line   between    OAk   and    OAk+1.      Let   a   denote  the 

angle  which  it  makes  with  OA.     Then  the  component  of  the 

attraction  at  0,  due  to  the  element  AkAk+l,  is  greater  than 

kp&y  cos  a  ,  ,        ,.        kp  Ay  cos « 

a  ,  /' — ,   A   x2  and  less  than  -^-^ — = — 

Let  cfik  denote  the  angle  AOAk.     Divide  the  inequality  by 

kp  Ay  cos  <f>k         ..  .     ,1      t    -i      T,    •  ti  ,i    , 

2    — 2 — >  an<^  Pass  ™  the  limit.     It   is   readily  seen  that 

the   limit   of    each   extreme   is   1.     Therefore   each   term   in 

the  component  attraction  of  AB  at  0  may  be  replaced  by 

kpAy  cos  <bk  kpcAy         .  c 

or  — £ — 2-j,  since  cos^  = 


x_  limit  yy  JkpcAy_ 


n  =  x> 


=  cfcp  J 


— -  =  — ,  ^        =  —  sin  ^10^. 

(c2  +  ^f     cVc2  +  62       c 


Similarly,   F=^f  ^ 


=kd 


^0(c2  +  2/2)f 

'    y<fy 

(c2  +  y2)i 
=  ^P(1- cos  AOB). 
kp 


Therefore  R  =  VX2  +  F2  =  -^  V2(l  -cos  .405) 

=  ^  sin  4.408, 

c  ^ 


334       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

and  its  line  of  action  makes  with  OA  an  angle  whose  tangent 

1S  Y     1-sinOBA     1-cosAOB      ,       -    ,-_ 

X        cos  OB  A  sm  AOB  2 

The  resultant  attraction  of  AB  at  0  therefore  bisects  the 
angle  AOB. 

ATTRACTION   OF  A  HOMOGENEOUS   CIRCULAR  DISC  AT  A 
POINT  IN   ITS   AXIS 

310.   Let  0  be  the  point,  and  ABC  the  given  circular  disc. 
Choose  the  center  of  the  disc  as  the  pole,  and  a  line  PA  in 
the  disc  as  the  initial  line. 

Let  c  denote  the  distance  of  0  from  the  pole,  and  a  the 
radius  of  the  disc.     Let  X  and  Y  denote  the  components  of 

the  attraction  of  the  disc  in  the 
direction  of  the  axis,  and  perpendic- 
ular to  the  axis  respectively.     Let  p 

denote  the  density  of  the  disc. 
i 

O        Divide    the    disc    up    into   polar 
elements  of  area  in  the  same  manner 
as  if  an  area  were  sought  by  double 
cV_y  '      '  integration.     Let  (rt,  6k)  be  the  coor- 

dinates of  the  corner  D  of  the  element  indicated.     (Fig.  125.) 

The  resultant  attraction  at  0  due  to  the  element  indicated  is 
greater  than  jcp(rArM  +  i* r*Aff) 


c8+(r,  +  Ar)2 

and  less  than  *^A/ +  f^>. 

cr  +  r. 

Let  the  resultant  attraction  of  this  element  make  an  angle  a 
with  PO.  Then  the  component  of  the  attraction  of  this  ele- 
ment in  the  direction  of  the  axis  is  greater  than 

fy)(r,ArA(9  +  ^Ar2Afl)  cos  a 
c2+(rJ  +  A?f 


ATTRACTION 


335 


j   i         i.i         kp(rLArAO  -f  {ArA^   cos  a      T  ,  , 

and  less  than  -£^ 9   2    9 * .     Let  <&ft  denote  the 

c2  +  rf 

i    -r,^-^     ^-   .-i    i,     •  t.     i      kprAr&O cos <bk       , 

angle  DOP.     Divide  the  inequality  by    r     2        2 — —  and  pass 

C    +  Tl 

to  limits.  The  limit  of  each  extreme,  as  A?*  and  A0  both 
approach  zero,  is  1.  Therefore  each  term  in  the  component 
attraction  of  the  disc  in  the  direction  of  the  axis  may  be  re- 


placed in  the  double  limit  by 

kpcrArAB     •  .  c 

-£ — - — — ,  since  cos  <j>k  =  - 

Therefore 


kprtAr A0  cos  <pk 
c2  +  ?f 


,  or  by 


X 


Vc2+n2 

= kpc£So 


=  2-n-kp 


1- 


2rr    rdrcW 

c 


Vc2  +  r2_ 

From  the  symmetry  of  the  figure,  or  by  direct  calculation,  it 
may  be  seen  that  Y  is  zero.     The  resultant  attraction  of  the 

and  its  line  of  action 


disc  at  0  is  therefore  2  irk J  1  — 


is  along  PO. 


L        Vc2  +  r2_ 


ATTRACTION  OF  A  HOMOGENEOUS   SPHERICAL   SHELL 

311.  Let  r0  be  the  inner  and  rx  the  outer  radius  of  the  shell. 
Let  P  be  the  center  of  the  shell  and  0,  distant  c  from  P, 
the  point  of  attraction. 
Through  OP  pass  a 
plane  OPH.  Divide  the 
area  of  the  circle  formed 
by  the  intersection  of 
the  sphere  and  the  plane 
up  into  infinitesimal 
elements  in  the  same 
manner  as  if  an  area 
were  sought  by  double 
integration.  Let  (r„  0k) 
be  the  coordinates  of  a  corner  of  the  element  of  area  indicated. 


Fig.  126. 


336       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

If  the  element  of  area  indicated  be  revolved  about  the  line 
OP,  the  volume  generated  is  such  that  the  limit  of  its  ratio  to 
2  Trty2  sin  0kArA0  as  A»*  and  A0  both  approach  zero  is  1.  (See 
Art.  217.)  Kevolve  the  element  of  area  through  an  angle  A<£. 
The  volume  generated  is  to  2  -n-r2  sin  0AAr  A0  as  A<£  is  to  2tt. 
The  volume  generated  by  revolving  the  area  through  an  angle 
A<£  may  then  be  taken  in  the  limit,  as  Ar  and  A0  both  approach 
zero,  as  r?  sin  dkArA$A<f>. 

The  attraction  at  0  of  the  mass  whose  volume  is  r*  sin  0AA?*A0A<£ 
is,  on  the  supposition  that  the  whole  mass  is  concentrated  at  D, 

kp?  am  0Ar±0±4>  where  f  =  OB2  =  c2  +  r2  -  2cr  cos  6k.      The 

y 

component  of  this  attraction  in  the  direction  of  PO  is  therefore 
kpr?  sin  flxArAflAcfr  cos  OPD 
f 
or  kpr?  sin  Qk(c  -  rt  cos  g^ArAgAj>?  ~. 

•  /^  riT^        C  —  Ti  COS  t/z. 

since  cos  OPD  = ? *• 

y 

It  is  evident  from  symmetry  that  the  component  of  this 
attraction  in  the  directions  PY  and  PZ,  perpendicular  to  PO 
and  each  other,  are  zero.  The  resultant  attraction  at  0  due  to  a 
mass  whose  volume  is  r?  sin  0AArA0A<£,  situated  at  D,  is  there- 
fore (1),  and  its  line  of  action  is  along  PO. 

It  can  be  shown  by  an  investigation,  presenting  no  difficulty 
other  than  that  of  length,  that  the  attraction  at  0,  due  to  the 
element  of  mass  whose  volume  is  indicated  in  the  figure,  lies 
between  two  extremes,  one  of  which  is  (1),  and  that  the  limit 
of  the  ratio  of  either  extreme  to  this  attraction  is  1.  Assum- 
ing that  this  limit  is  1,  we  have,  from  the  theorem  of  Art.  186, 
that  the  attraction  of  the  element  of  mass  may  be  taken  as  (1) 
in  the  limit. 

The  attraction  at  0  of  the  whole  spherical  shell  is  therefore 

rrx  C*  r2n  r2  (c  —  r  cos  6)  sin  $  dr  dO  d<f> 

PJrQ    J)     Jo  f 


ATTRACTION  837 

Substitute  for  0  its  value  in  terms  of  y.     Therefore  the 
attraction         ,       _      _     „_     ,  9       „  ,     2x 

_i   p  p  p  K^-^  +  ^tfrdy^ 

2  C2  Jr0    JyQ  Jo  if 

c2  •/•■o      L         2/         J  I* 

To  determine  ?/0  an(i  Vu  we  must  distinguish  between  two 
cases. 

Case  I.     Where  0  is  a  point  inside  the  shell. 
Then  y0  =  r  —  c,  and  ?/!  =  ?'  +  c. 

Therefore  the  attraction  =  ^  f1  0  dr  =0 . 

C2    Jr0 

Case  II.     Where  0  is  a  point  outside  the  shell. 
Then  y0  =  c  —  r,  and  y1=c  +  r. 

Therefore  the  attraction  =    """  p  |   *  i^dr 

C2      Jr„ 


=  ^f(n3-V).  (2) 


If  in  (2)  we  make  r0=0,  we  have  the  attraction  of  a  solid 
sphere  of  radius  rx  and  density  p  at  a  point  outside  the  sphere 
and  distant  c  from  the  center  to  be 

4  7rlcp1\3 


3  c2 


EXERCISES 


1.  Show  that  the  value  of  Jc  in  the  F.P.S.  absolute  system 
equals  1033  x  10"12.     (See  Art.  306.) 

2.  Find  the  attraction  due  to  a  homogeneous  straight  wire 
of  length  2 1  at  a  point  in  the  line  of  the  wire  and  distant  c 

from  one- end.  A  2kpl_  .      h   direction of  the  wire. 

c(c+2Z) 


338       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

3.  Find  the  attraction  due  to  a  homogeneous  straight  wire 
of  length  2  I  at  a  point  distant  c  from  the  center  of  the  wire 
and  on  a  line  perpendicular  to  the  wire. 

Ans.  . P     ,  in  the  direction  of  the  line  perpendicular  to 

the  wire. 

4.  "Find  the  attraction  due  to  a  homogeneous  right  circular 
cylinder  of  length  2 1  and  radius  of  cross  section  a,  at  a  point 
in  the  axis  produced  of  the  cylinder  and  distant  c  from  one  end. 

Ans.  2 Trkp [2 1  +  VaF+c2 -  V(c  +  2 If  +  a2],  in  the  direction 
of  the  axis  of  the  cylinder. 

5.  From  the  result  in  Exercise  4,  show  that  the  attraction 
of  a  homogeneous  right  circular  cylinder  of  radius  of  cross 
section  a,  and  infinite  in  one  direction,  at  the  point  of  inter- 
section of  the  axis  and  base,  is  2  -n-kpa. 

6.  Three  rods,  equal  in  length  and  of  uniform  density,  form 
a  triangle.  Find  the  point  at  which  the  resultant  attraction  is 
zero.  Ans.   The  center  of  the  inscribed  circle. 

7.  Find  the  attraction  of  a  homogeneous  straight  wire  of 
infinite  length  at  a  point  c  units  distant  from  it. 

Ans.    — £,  in  the  direction  perpendicular  to  the  wire. 

8.  Find  the  attraction  of  a  homogeneous  right  circular  cone 
of  vertical  angle  2  a  and  height  h,  at  the  vertex  of  the  cone. 

Ans.   2  7rJcp(l  —  cos  a)h,  in  the  direction  of  the  axis. 

9.  Find  the  attraction  of  a  homogeneous  right  circular 
C}rlinder  of  iron,  10  meters  long,  radius  of  cross  section  1  meter, 
upon  a  mass  of  1  kilogram  in  the  axis  of  the  cylinder,  distant 
10  cm.  from  one  end.    (Weight  of  iron  =  7.23  grams  per  cu.  cm.) 

Ans.  0.000256  gram,  in  the  direction  of  the  axis  of  the 
cylinder. 

10.  Supposing  the  earth  at  rest  and  the  resistance  of  the  air 
zero,  with  what  velocity  must  a  body  be  projected  vertically 
upwards  from  the  surface  of  the  earth  in  order  that  it  may 
never  return  ?  (See  Art.  278.  Assume  radius  of  earth  = 
4000  miles.)  Ans.   36800  ft.  per  sec.  (nearly). 


CHAPTER   XXXVI 

OENTEKS   OF   GKAYITY 

Definition.  Two  parallel  forces  are  said  to  be  like  or  unlike 
according  as  they  act  in  the  same  direction  or  in  opposite 
directions. 

312.  Suppose  that  there  are  two  like  parallel  forces  acting 
at  points  in  a  rigid  body.  To  determine  the  resultant  of  the 
forces. 

Let  the  forces  be  P1  and  P2,  acting  at  the  points  A  and  B 
respectively  (see  Fig.  127). 

At  A  introduce  a  force  F  acting  in  the  direction  BA,  and  at 
B  an  equal  force  F  acting  in  the  direction  AB.  Complete  the 
parallelograms  AC  and  BD. 

F        G      F 


Fig.  127. 


The  forces  F  and  Px  at  A,  and  the  forces  F  and  P2  at  B  are 
equivalent  to  two  forces  represented  in  magnitude  and  direction 
by  AC  and  BD  respectively,  and,  since  the  forces  F  neutralize 

339 


340       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

each  other,  the  forces  Px  and  P2  are  therefore  equivalent  to 
these  forces. 

Produce  CA  and  DB  to  meet  in  G.  At  G  resolve  the  forces 
represented  by  AC  and  BD  into  forces  equal  and  parallel  to 
the  original  ones,  F,  Plf  and  P2. 

Since  the  forces  Pat  G  neutralize  each  other,  the  resultant  of 
the  forces  acting  at  A  and  B  is  equivalent  to  the  force  Px  +  P2 
acting  at  G. 

To  determine  the  point  at  which  the  line  of  action  of  the  re- 
sultant force  cuts  the  line  AB : 

The  triangles  G MA  and  AHG  are  similar. 

GM=  AH=  P1 
MA     HC      F' 

The  triangles  GMB  and  BKD  are  similar. 

GM     BK  _    P2 


Therefore,  by  division, 


MB 

KD      -F 

MB_      P1 

MA         P2 

.-.   -Px 

.  MA  =  P2>  MB. 

.-.  P, 

•  AM=  P2  •  MB. 

The  resultant  of  two  like  parallel  forces  P1  and  P2,  acting  at 

points  A  and  B  in  a  rigid  body,  is  therefore  a  force  P1  +  P2 

parallel  to  Px  and  P2,  acting  at  a  point  M  on  the  line  joining 

A  and  P,  such  that 

Px  •  ^Of  =  P2  •  MB. 

Since  Px  •  AM  =  P2  •  iHfP,  whatever  be  the  direction  of  the 
forces  Px  and  P2,  it  follows  that  the  same  point  M  would  be 
found  if  P1  and  P2  were  revolved  about  their  points  of  applica- 
tion, remaining  parallel. 


CENTERS   OF  GRAVITY 


341 


313.  Given  the  coordinates  of  the  points  A  and  B  at  which 
the  forces  P1  and  P2  act,  we  can  find  the  coordinates  of  the 
point  31,  through  which  their  resultant  acts,  as  follows: 

Let  the  coordinates  of  A,  B,  and  M  be  (x1}  yl9  zx),  (x2,  y2,  z2), 
and  (x,  y,  z)  respectively. 


Fig.  128. 
Since  P1  •  AM=  P2  •  MB, 

•     •        -*1\  *^  *^1/     ""^    -^~2\        2   vOjm 

.'.   (P1  +  P2)x  =  P1x1  +  P2x2. 

-L\X]_  ~\-  -L2X2 


X. 


Pl+P2 


Similarly, 
and 


P^  +  P^ 
Pi  +  P2 


314.  Suppose  that  there  are  n  like  parallel  forces,  Plt  P2, 
P3,  ••-,  Pn,  acting  at  points  Alf  A2,  As,  ...,  An  respectively,  in  a 
rigid  body.     To  determine  the  resultant  of  the  forces. 

Suppose  that  the  points  of  application  of  the  forces  are 
given  by  their  coordinates.     Let  (aj15  yx,  zx),  (x2,  y2,  z2),  (x3,  ys,  z3), 


342       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

*••>  (xn>  Vn,  »n)  be  the  coordinates  of  the  points  A1}  A2,  A3,    ••, 
An  respectively. 


Fig.  129. 

By  Art.  313,  the  resultant  of  P1  and  P2  is  a  force  Px  +  P2 
acting  at  a  point  Jf1?  coordinates  (x',  y',  z'),  such  that 

Pi(y'-yi)=  Ply*  -y')> 

PjQi'  -  zx)  =  P2(z2  -  z'). 

.  t  =  PlXl  +  P2X2       yl  _  A.Vl  +  P2?/2        „f  =  A^l  +  P2% 

-Pi  +  p2  '  *       Pi  +  p2  '  "       A  +  p2 

Similarly,  the  resultant  of  Px  +  P2,  acting  at  jM^  and  P3  at 
u43,  is  a  force  Px  +  P2  +  P3,  acting  at  a  point  Jf2?  coordinates 
(a;",  2/",  z"),  such  that 

(A  +  A)0"  -  «')  =  P3fe  -  x"), 

(-Pi  +  P2)(2/"-2/')  =  A(?/3-2/,,)J 
(P1  +  P2)(^-^)  =  P3(%-^). 

„  (Pt  +  P2>'  +  P3%  =  -Pl^l  +  P2^2  +  P3^3 

'"'   "  P1  +  P2  +  A  Pl-fP2  +  P3         ' 


y"  = 


(Pi  +  p2W  +  A^/s  _  A?/i  +  P*h  +  ^3 


px  +  p2  +  p3 


px  +  p2  +  p3 


„  _  (P,   +  P»>'  +  A*3  _  A*l  +  P^2  +  P3% 
Px  +  P2  +  P3  -Pi  +  P'2  +  Ps 


CENTERS   OF  GRAVITY  343 

By  continuing  this  process,  we  find  finally  that  the  resultant 
of  the  n  like  parallel  forces  is  a  force  Pl  +  P2  -f  P3  -f  •••  +  Pn, 
acting  at  a  point  Mn_1}  coordinates  (x,  y,  z),  such  that 

_  Pyx,  +  p&  +  P&  +  •  •  •  +  Pa 


aj 


A  +  ^  +  A+-+^n  ' 


5  _^  fi//l  +  A?/2  +  P,?h  +   "   +  P^, 

=  Plgl  +  P2*2  +  P3Z3  +   ..■   +  PMg„ 
Pl  +  P2  +  P3+-+P« 

Since  the  expressions  for  x,  y,  and  z  are  independent  of  the 
direction  of  the  forces,  it  follows  that  the  same  point  Mn_x  will 
be  found  if  the  forces  be  revolved  about  their  points  of  applica- 
tion remaining  parallel.  Also,  since  a  variation  of  the  order  in 
which  the  forces  were  combined  would  cause  only  a  variation  in 
the  order  of  the  terms  in  x,  y,  and  z,  it  follows  that  the  same 
point  Mn_x  will  be  determined  if  the  forces  be  combined  in  any 
other  order. 

Definition.  The  point  Mn_ly  coordinates  (x,  y,  z),  is  called  the 
center  of  the  system  of  parallel  forces. 

315.  An  important  special  case  of  like  parallel  forces  acting 
at  points  in  a  rigid  body  is  that  of  gravity  acting  on  n  particles 
rigidly  connected. 

Let  m^  m2,  m3,  •••,  mn  be  the  masses  of  the  particles.  Then, 
by  the  preceding  article,  the  coordinates  of  the  center  of  the 
system  of  these  parallel  forces  acting  on  these  masses  are 

-  _  m1gx1  +  m2gx2  +  m3gx?>  +  ••-  -f  mnqxn 

jj  — — ; 

mxg  +  m2g  +  m<g -\ \-  mng 

_  mxxx  +  m2x2  +  m8a?3  +  •  •  •  +  m>nxn 
my  +  m2  +  m3  -f  •••mn 


344       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

-  _  m1gy1  +  m2gy2  +  msgy3  -j h  m,uqyn 

mxg  +  m2g  +  m3g  H 1-  mng 


_  mhy1  +  m2y2  +  m3y3  + \-  m^yrt 

mx  +  m2  +  m3  +  ...  +  mn 

z  =  m^gj  +  m2gz2  +  m3^3  j h  m^, 

m^  +  m2g  +  m^  H \-  m„g 

_  m^x  +  m2z2  +  m3g3  H h  mw^w 

mx  +m2  +  m3H \-  mn 

Definition.  The  center  of  the  system  of  parallel  forces  of 
gravity  acting  on  a  system  of  particles  rigidly  connected  is 
called  the  center  of  gravity  of  the  system  of  particles. 

Center  of  gravity  is  usually  denoted  by  the  letters  C.G-. 

316.  Let  a  force  P  have  its  line  of  action  FB  oblique  to  the 

line  AG  (see  Fig.  130). 

At  A  draw  a  plane  perpen- 
dicular to  AC,  cutting  FB  in  F. 
Let  FH  be  the  projection  of 
FB  on  this  plane.  From  A 
draw  a  perpendicular  to  FH. 

Denote  the  angle  HFB  by  <£, 
and  the  length  of  the  perpen- 
Fig.  130.  dicular  from  A  to  FH  by  a. 

Definition  of  moment.  The  moment  of  the  force  P  about  the 
line  AC  is  defined  to  be  aPcos  <£. 

317.  From  the  definition  of  moment,  it  follows  that : 

First:  If  the  line  of  action  of  P  is  perpendicular  to  AC, 
then  $  =  0,  and  therefore  the  moment  of  P  about  AC  is  the 
product  of  the  magnitude  of  P  and  the  perpendicular  distance 
of  the  line  of  action  of  P  from  AC. 

Second:  If  AC  and  the  line  of  action  of  P  are  in  the  same 
plane  and  not  parallel,  a  is  zero,  and  therefore  the  moment  of 
P  about  AC  is  zero. 


CENTERS   OF  GRAVITY  345 

If  AC  and  the  line  of  action  of  P  are  parallel,  <£  =  90°,  and 
therefore  the  moment  of  P  about  AC  is  zero. 

318.  The  equations  of  Art.  315  may  be  written  in  the  form 

x  .  2m  =  2m#, 

y  -  2m  =  2m?/, 

z  •  2m  =  2mz. 

From  these  equations  we  see  that  the  moment  of  the  sum  of 
the  masses,  considered  as  concentrated  at  the  C.G.,  about  any- 
one of  the  coordinate  axes,  is  equal  to  the  sum  of  the  moments 
of  the  masses  about  that  axis. 

319.  In  the  case  of  a  continuous  body  bounded  by  known 
surfaces,  the  coordinates  of  the  C.Gr.,  (x,  y,  z),  may  be  found 
approximately  by  dividing  the  body  up  into  infinitesimal 
elements  of  mass  Am  and  determining  x,  y,  and  z  so  that 

-      1x  Am 
x  = ? 

2  Am 

1y  Am 

y  =  — > 

2  Am 

1z  Am 
z= , 

2  Am 

where  the  summation  includes  all  the  elements  of  mass,  Am, 
and  x,  y,  and  z  are  the  distances  from  the  axes  of  some  point 
in  Am. 

Definition.  The  limits  which  x,  y,  and  z  approach  as  Am 
approaches  zero  determine  the  coordinates  of  the  C.G-.  of  the 

body.     Thus,      r  C  C 

I  x  dm  I  y  dm  J  z  dm 

x  =  ^- ,     y  =  - ,      z  =  ^ 


dm  I  dm  I  dm 


where  the  limits  of  integration  are  such  as  to  embrace  the 
whole  solid. 


346       DIFFERENTIAL  xiNB  INTEGRAL   CALCULUS 


320.   The  above  equations  may  be  written 

x  I  dm  =  I  x  dm, 

yjdm=  jydm, 

z  I  dm  =  I  z  dm. 


From  these  equations  we  see  that  the  moment  of  the  body 
considered  as  concentrated  at  C.G.,  about  any  one  of  the  co- 
ordinate axes,  is  equal  to  the  limit  of  the  sum  of  the  moments 
of  the  elements  about  that  axis. 

321.  Find  the  coordinates  of  the  C.G-.  of  a  homogeneous 
plane  area  when  the  equation  of  the  bounding  curve  is  given 
in  rectangular  coordinates. 

Suppose  that  the  bounding  curve  is  as  in  Fig.  131. 

Divide  the  area  up  into  rec- 
tangular elements  of  area  Ax  Ay 
in  the  same  manner  as  if  an 
area  by  double  integration  were 
sought. 

Let  (x,  y)  be  the  coordinates  of 

_     the  C.G.     The  mass  of  each  ele- 

ment  is  p  Ax  Ay,  where  p  is  the 

mass  per  unit  of  area.     Take  an 

element  ABCD.     Suppose  that  the  coordinates  of  A  are  (a?w  yt). 

Then  the  coordinates  of  B  are  (xk-\-  Ax,  ?/,). 

The  moment  of  the  element  ABCD  about  the  y-axis  is  evi- 
dently greater  than  if  the  whole  mass  were  concentrated  at 
A  and  less  than  if  it  were  concentrated  at  B. 

A     A        moment  of  ABCD   .  ,      ,    .    N     .     A 
■•*A,4»<    about  the  2/-axis    <  <*  +  Ax>  Ax  ^ 

Divide  the  inequality  by  x,.p  Ax  Ay  and  pass  to  the  limit. 
Since  the  limit  of  each  extreme  is  1, 


Fig.  131. 


limit 


'moment  of  ABCD  about  the  ?/-axis" 
xL.p  Ax  Ay 


=  1. 


CENTERS  OF  GRAVITY 


347 


x  = 


Similarly, 


I   J  px  dx  dy 

I  p  dx  dy 

\\ptydxdy 

I    I  p  dx  dy 


In  these  integrals  the  limits  are  the  same  as  if  an  area  were 
sought. 

Since  p  is  constant  it  may  be  removed  from  under  the  inte- 
gral signs  and  cancelled. 

I  x  dx  dy  I    \  y  dx  dy 

Therefore    x  =  ^— ,  and  y  =  ^- 

dx  dy  dx  dy 

Since   I   I  dx  dy  denotes  the  area,  we  have 

I  xdxdy  I    I  y  dxdy 


;/:  = 


area 


area 


322.  Find  the  coordinates  of  the  C.G-.  of  a  homogeneous 
plane  area  when  the  equation  of  the  bounding  curve  is  given 
in  polar  coordinates. 

Suppose  that  the  curve  is 
as  in  Fig.  132. 

Divide    the    area    up    into 
polar  elements  of  area 

r\0  Ar  +  ^-A^Afl 

in  the  same  manner  as  if  an 
area  were  sought. 

Let  the  coordinates  of  the  C.G-.  be  (x,  y),  where  x  is  the  dis- 
tance of  the  point  from  a  line  OB  perpendicular  to  the  initial 
line  OA,  and  y  is  the  distance  of  the  point  from  OA. 

The  mass  of  each  element  is  p(r\9  Ar  +  \  Ar2  A0j. 


Fig.  132. 


348       DIFFERENTIAL   AND  INTEGRAL   CALCULUS 

Take  an  element  CDEF.  Suppose  that  the  coordinates  of 
C  are  (r?,  6k).  Then  the  coordinates  of  D  are  (rt  +  Ar,  0k)  and 
the  coordinates  of  F  are  (rt,  0k  -f  A0). 

The  moment  of  CDEF  about  05  is  evidently  greater  than 
if  the  whole  mass  were  concentrated  at  F  and  less  than  if 
it  were  concentrated  at  D. 

.-.  rt P  cos  (Ok  +  A0) (r,  A#  Ar  +  J  Ar2  A0)  <  moment  of  CDEF 

about  05  <  (r,  +  Ar)p  cos  ft.(r,  A0  Ar  + 1  A?-2  A0). 

Divide  the  inequality  by  rtp  cos  0^  A0  Ar)  or  r*p  cos  0^  A0  Ar, 
and  pass  to  the  limit.     Since  the  limit  of  each  extreme  is  1, 


limit 


moment  of  CD  EF  about  OB 
rfp  cos  6k  A0  Ar 

I    j  i2p  cos  0  d0  dr 


=  1. 


x  = 


I    |  rp  dO  dr 

I    I  r2/o  sin  0  d0  dr 

j   j  rp  dO  dr 

Or,  as  in  the  preceding  article, 


Similarly, 


J    j  r2  cos  0  d$  dr  j   I  r2  sin  0  dO  dr 


x  =  ^—^- ,  and  y  = 


area  area 


323.  By  methods  entirely  similar  to  those  of  Arts.  321  and 
322,  we  can  find  that  the  coordinates  of  the  C.G-. : 

Of  a  homogeneous  arc  when  the  equation  of  the  bounding 
curve  is  given  in  rectangular  coordinates  are : 


x  = 


s 


ds 


(yds 

y=-7= — 

jds 


CENTERS   OF  GRAVITY 


349 


Of  a  homogeneous  arc  when  the  equation  of  the  bounding 
curve  is  given  in  polar  coordinates  are : 


I  r cos  Oris  I  rs 

x=^— ■= ,      y=*L—/ 


sin  6  ds 


Jf* 


fds 


Of  a  homogeneous  volume  of  revolution  when  the  equation 
of  the  generating  arc  is  given  in  rectangular  coordinates  are : 


x  = 


ss 


xy  dx  dy 


j  jydxdy 


Jf=0. 

Of  a  homogeneous  volume  of  revolution  when  the  equation 
of  the  generating  arc  is  given  in  polar  coordinates  are : 


.  // 


?-3  sin  0  cos  6  d$  dr 


x  = 


y  =  0. 


SS* si 


sin  6  dO  dr 


Of  a  homogeneous  surface  of  revolution  when  the  equation 
of  the  generating  arc  is  given  in  rectangular  coordinates  are : 


I  xyds 

J  yds 


y  =  0. 

324.    To  find  the  coordinates  of     Y 
the  CO.  of  a  given  homogeneous 
plane  triangle. 

Let  the   triangle  and  its   coor- 
dinate axes  be  as  in  Fig.  133. 

Let  '(x,  y)  be  the  coordinates  of 
the  C.G. 


Fig.  133. 


350       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


The  equations  of  the  lines  OB  and  PB  are 

1 


y  =  -x,  and  y  = (a  —  x)  respectively. 


T"  7(0-— x) 

f*b    /•  b  f*a    f*a—b 

I     I     x  dx  dy  +  I     I  a;  dx  dy 

- _JoJo Jb  Jo 

X  —  7,     : 


area  of  A 


a    fa—b 


and 


ydxdy-\-  I     I 

Jb   Jo 


(a—x) 


ydxdy 


area  of  A 


a  +  b 


It    will    be    remembered    from     analytic     geometry    that 

CL    I     7)      c\ 

,  -  )  are  the  coordinates  of  the  point  of  intersection  of 

the  medians  of  the  triangle.     The  C.G.  of  a  plane  triangle  is 
therefore  the  point  of  intersection  of  the  medians. 

325.  If  a  given  area  is  the  sum  or  the  difference  of  the  areas 
of  figures  whose  areas  and  centers  of  gravity  are  known,  the 
C.G.  of  the  given  area  can  be  found  without  resort  to  inte- 
gration. 

Example.  The  side  00  of  the  homogeneous  square  OBDG 
(Fig.  134)  is  bisected  in  A,  and  the  triangle  OBA  is  turned 

about  AB  to  the  position  OBC.     To 
find  the  C.G.  of  the  shaded  area. 

Let  the  sides  of  the  square  be  2  a. 
Take  the  coordinate  axes  as  indi- 
cated. 

The  C.G.  of  the  square  is  evidently 
at  the  point  whose  coordinates  are 
(a,  a).     Let  (x,  y)  be  the  coordinates 
of  the  C.G.  of  the  shaded  area.     The 
area  of  the  shaded  figure  =  4  a2.     The  areas  of  the  triangles 

ra 


OBC  and  OB  A  are  each  =  a2. 


and  of  OBC  is  (-%  ^\     (See  Art.  324.) 


u         O  J 


CENTERS  OF  GRAVITY  351 

The  moment  of  the  shaded  figure  about  the  ?/-axis  =  moment 
of  square,  minus  moment  of  triangle  OBA,  minus  moment  of 
triangle  OBC,  each  about  the  ?/-axis. 

a2  •  x  =  4  a2  •  a  —  a2  •  -  —  <r  •  -• 

o  o 

Similarly,  ?/  =  a,  a  result  which  would  be  expected  from  the 
symmetry  of  the  figure. 

THEOREMS   OF  PAPPUS 

326.  Theorem  I.  The  area  of  the  surface  generated  by  the 
revolution  of  an  arc  of  a  plane  curve  about  an  axis  in  its  plane 
and  not  crossing  the  arc  is  equal  to  the  product  of  the  length 
of  the  arc  and  the  length  of  the  path  described  by  the  C.G.  of 
the  arc. 

Proof.  Let  S  denote  the  area  of  the  surface,  s  the  length 
of  the  arc,  and  y  the  distance  of  the  C.G.  of  the  arc  from  the 
line.  Cy  ds 

ThenS=2>7rCyds.    (See  Art. 200.)    Also,  y=^ (See 

Art.  323.)  o  o   - 

/  .-.  8  =  s  •  2  iry. 

Theorem  II.  The  volume  generated  by  the  revolution  of  a 
plane  area  about  an  axis  in  its  plane  and  not  crossing  the  area 
is  equal  to  the  product  of  the  area  and  the  length  of  the  path 
described  by  the  C.G.  of  the  area. 

Proof.     Let  V  denote   the   volume,  A  the  area,  and  y  the 

distance  of  the  C.G.  of  the  area  from  the  line.  r 

/\\  y dx 
y2dx.   (See  Art.  204.)  Also,  y  =  -^— (See 

Art.  323.)  jr      a    9    - 

We  have  thus  far  supposed  the  bodies  considered  to  be 
homogeneous.  When  they  are  non-homogeneous,  the  formulas 
for  C.G.'  differ  from  those  given  above  only  by  the  presence  of 
a  factor  p  in  each  integrand. 


352       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

EXERCISES 

1.  Express,  by  single  integrals,  the  coordinates  of  the  C.G. 
of  a  homogeneous  plane  area  when  the  equation  of  the  bound- 
ing curve  is  given  in  rectangular  coordinates. 

2.  Find  the  coordinates  of  the  C.G.  of  a  segment  of  the 
parabola  y2  =  4  ax  cut  off  by  the  chord  x  =  h. 

Ans.  x  =  ^h,  y  =  0. 

3.  Find  the  coordinates  of  the  C.G.  of  the  area  of  a  semi- 
circle of  radius  a.  j^ns   %  _  £^   ^  =  0. 

4.  Find  the  coordinates  of  the  C.G.  of  a  semi-ellipse,  semi- 
axes  a  and  b,  the  bisecting  line  being  the  major  axis. 

4  6 
Ans.  x  =  — ,  y  —  0. 

3  7T 

5.  Find   the   coordinates   of   the   C.G.    of   the    plane   area 

2  2  2  • 

bounded  by  the  curve  xs  -f  y*  =  a*,  which  lies  in  the  first  quad- 
rant. A       -     -     256  a 

Ans.  x=y  = . 

9         315  7T 

6.  Find  the  coordinates  of  the  C.G.  of  the  cycloid 

x  =  a(&  —  sin  0),  y  =  a(l  —  cos  6). 

Ans.    x  =  air,  y  =  ^a. 

7.  Find   the  coordinates  of   the  C.G.  of  one  loop  of   the 

lemniscate  r2  =  a2  cos  2  6.  '     _     v  -x/2      - 

Ans.   x  = a,  y  =  0. 

8 

8.  Find  the  coordinates  of  the  C.G.  of  a  circular  sector  of 
angle  2  a. 

Ans.   If  the  radius  bisecting  the  sector  is  on  the  a>axis, 

-      2  a  sin  a    —     A 
o       a 

9.  Find  the  coordinates  of  the  C.G.  of  the  cardioide 
r  =  a(l  +  cos  6)  on  the  right  side  of  the  line  through  the  pole 

perpendicular  to  the  initial  line. 

A       -     16  +  5tt      -        10  a 
Ans.  x  =  —  a,  y  = 


16  +  6' 7T   ;  °      8  +  3 


CENTERS   OF  GRAVITY  353 

10.  Find  the  coordinates  of  the  C.G.  of  the  arc  of  the  curve 

2  2  2  

x*  -f-  y3 '  =  a's  in  the  first  quadrant.  Ans.    x  =  y  =  \a. 

11.  Find  the  coordinates  of  the  C.G.  of  an  arc  of  a  semi- 
cycloid,  the  equations  of  the  cycloid  being  x  =  a{6  —  sin  6), 
y  =  a(l  —  cos  0).  Ans.   x  =  (71- —  J)a,  2/  =  —  -|  a. 

12.  Find  the  coordinates  of  the  C.G.  of  the  solid  formed  by 
revolving  the  sector  of  a  circle,  angle  2  a,  about  one  of  its 
extreme  radii. 

Ans.  If  an  extreme  radius  is  on  the  a>axis,  ~x  —  \a  cos2  a,  y  =  0. 

13.  Find  the  coordinates  of  the  C.G.  of  the  segment  of  the 
paraboloid,  formed  by  revolving  the  parabola  ?/2=4  ax  about  the 
ce-axis,  cut  off  by  the  plane  x  =  li.  Ans.   x  =  ^h,  y  =  0. 

14.  Find  the  coordinates  of  the  C.G.  of  the  surface  formed 
by  revolving  the  cardioide  r=a(l  +  cos0)  about  the  initial  line. 

Ans.    x  =  ^a,  y  =  0. 

15.  Find  the  coordinates  of  the  C.G.  of  a  hemispherical 
surface.  A        fa 


Ans.   (!<). 


16.  Find  the  coordinates  of  the  C.G.  of  a  hemisphere  whose 
density  varies  as  the  distance  from  the  center  of  the  sphere. 

Ans.    (|  a,  0.) 

17.  Find  the  distance  between  the  center  and  the  C.G.  of 

one  half  an  anchor  ring  generated  by  a  circle  of  radius  a  whose 

center  describes  a  circle  of  radius  b.  a       ±  b2  -4-  a2 

A.ns. —  • 

2tt6 

18.  Find  the  coordinates  of  the  C.G.  of  a  volume  when  the 
equation   of    the   bounding   surface   is    given   in   rectangular 

I   I   I  xdxdydz            J    J    J  ydxdy dz 
Ans.   x—J  J  J y=J  J  J 

vol.  '*  vol.  ' 


J    I    \zdxdy  dz 
vol. 


2a 


354       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

19.  A  cone  of  height  h  is  cut  out  of  a  cylinder  of  the  same 
base  and  height.  Find  the  distance  of  the  C.G.  of  the 
remainder  from  the  vertex.  Ans.   -I  h. 


20.  Find  the  coordinates  of  the  C.G.  of  the  surface  of  a 
solid  of  revolution  when  the  equation  of  the  bounding  curve  is 
given  in  polar  coordinates.  ^       ft*  cob  9  An  6  da    _ 

Ans.   x—^- — ,  2/  =  0. 


/• 


r  sin  9  ds 


21.   Prove  that  the  x  of  the  C.G.  of  a  set  of  n  plane  areas  in 
the  same  plane  is 

-_«^i  +  x2F2  +  x3F3+.~+xnFn 


where  xk  is  the  x  of  the  C.G.  of  the  area  Fk. 

22.   Find  the  coordinates  of  the  C.G.  of  the  area  between  a 
quadrant  of  a  circle  and  the  circumscribing  square. 


Ans.   x  =  y  = 


2a 


3(4 -tt) 


Find  the  position  of  the  C.G.  of  each  of  the  following  areas: 


23. 

. 

l 

~7 

> 

l 

— > 

1 

v — 

— ^ > 

I*. — 


-r- 


~*!' 


24. 


23.    ^.ns.  a; 


-     a2  +  at-t2 


2(2a-t) 


.A 


«L_ 


L 


24.    J.ns.  a;  =  2.718". 


25. 


CENTERS   OF   GRAVITY 

26, 


355 


kt> 


2  ft  -  4  *  +  4  6 


27. 


26.    ^4ws.  #  =  0.638". 


^L_ 


27.   -4tis.  x  =  2.568". 

Find,  by  Pappus'  Theorem  : 

28.  The  surface  and  volume  of  a  sphere. 

29.  The  surface  and  volume  of  the  torus  generated  by  the 
curve  x2  +  (y  —  b)2  =  a2,b>  a. 

30.  The  surface  and  volume  of  the  solid  formed  by  revolving 
the  cycloid  about  its  base. 


CHAPTER   XXXVII 

MOMENTS   OP  INEETIA 

327.  Definition.  The  moment  of  inertia  of  a  particle  with 
reference  to  a  point,  line,  or  plane  is  the  product  of  the  mass  of 
the  particle  and  the  square  of  the  distance  of  the  particle  from 
the  point,  line,  or  plane. 

Thus,  if  m  be  the  mass  of  a  particle  and  r  the  distance  of 
the  particle  from  a  point,  line,  or  plane,  the  moment  of  inertia 
of  the  particle  with  reference  to  the  point,  line,  or  plane  is  rar2. 

328.  Suppose  that  there  are  n  particles  of  masses  m1}  m2,  m3, 
•  ••,  mn  situated  at  points  Al9  A2,  A3,  •••,  ^respectively.  To 
determine  the  moment  of  inertia  of  the  system  with  reference 
to  the  coordinate  axes. 

Let  the  coordinates  of  Al9  A2,  A3,  ■••,  An  be  (x±,  ylt  %),  (x2,  y2, 
z2),  fa,  Vs,  %),  — ,  0»«  Vn,  «»)  respectively.  Let  the  moments 
of  inertia  of  the  system  with  reference  to  the  a-,  y,  and  z  axes 
be  denoted  by  Ix,  Iy,  Iz  respectively. 

Then 

-4  =  WiO/i2  +  Zi2)  +  m2(y22  +  z22)  +  m3(y32  +  z32)  +  •  ■  •  +  mn(yn2  +  O 

=  ^m(y2  +  z2). 

Iy  =  miOi2  +  %*)  +  m2(a;22  +  z})  +  m3(x32  +  %2)  +  —  +  mn(xn2  +  < 

=  Xm(a2  -f-  z2). 

Iz  =  mi(x2  +  ?/i2)  +  m2(x2  +  2/22)  +  m3(x2  +  2/32)  +  •  ••  +  mn(x-n+  yl) 

=  ^m(ar  +  2/2)- 
356 


MOMENTS   OF  INERTIA  357 

329.  The  moment  of  inertia  of  a  continuous  body  with 
reference  to  a  point,  line,  or  plane  can  be  found  approximately 
by  dividing  the  body  up  into  infinitesimal  elements  of  mass, 
A»i,  forming  the  product  of  each  element  of  mass  and  the 
square  of  the  distance  of  some  point  in  it  from  the  given  point, 
line,  or  plane,  and  taking  the  sum  of  the  results. 

Definition.  The  limit  which  this  sum  approaches  as  Am  ap- 
proaches zero  is  the  moment  of  inertia  of  the  body  with  refer- 
ence to  the  point,  line,  or  plane. 

Thus,  the  moments  of  inertia  of  a  continuous  body  with 
reference  to  the  coordinate  axes  are,  approximately : 

where  x,  y,  and  z  are  the  distances  of  some  point  in  the  ele- 
ments of  mass,  Am,  from  the  x,  y,  and  z-axis  respectively,  and 
the  summation  includes  all  the  elements  of  mass  Am.  The 
moments  of  inertia  of  the  body  with  reference  to  the  axes 

are :  ^ 

=  J  (y2  +  ^)dm, 


L 


=  J  {x-  +  z2)dm, 
=  I  (x'2  +  y2)dm, 


where  the  limits  embrace  the  whole  solid. 

•  330.  As  illustrations  of  the  method  of  determining  the  mo- 
ment of  inertia  of  a  continuous  body,  consider  the  following 
examples : 

Example  1.  Find  the  moment  of  inertia  of  a  homogeneous 
rod  of- length  I  and  density  p  with  reference  to  an  axis  perpen- 
dicular to  the  rod  through  one  end. 


358       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


Let  AB  (Fig.  135)  be  the  given  rod,  and  A  the  end  through 

j  ;  ,    which  the  axis  passes. 

A  Ak^k-i-i        0        Divide  AB  into  n  equal  parts. 

Fig.  135.  CaU  each  part  A^ 

Let  AkAk+1  be  a  length  Ax  where  Ak  is  distant  xk  from  A 

3  =  1 

Therefore  the  required  moment  of  inertia  =  ll™it  X  tfpAx. 


->i 


x=0 


a&fcc. 


=  P 


Z3 


If  Jf  is  the  mass  of  the  rod,  the  required  moment  of  inertia 

3 

Example  2.     Find  the  moment  of  inertia  of  a  homogeneous 
right  circular  cone  of  height  h  and 
radius  of  base  a  with  reference  to 
its  axis. 

Divide  the  volume  of  the  cone 
up  into  infinitesimal  elements  in 
the  same  manner  as  if  a  volume 
were  sought  by  double  integration. 
(See  Art.  214.) 

The  element  of  volume 


Fig.  136. 


=  2  7r  [i/i  Ay  Ax  +  i  A?/2  Ax]. 
Therefore  the  required  moment  of  inertia 


x=h 


_  limit  V1 


x=0 


— -X 

h 


y= 


I  o        limit  V^ 

\27rPm=^2^y3Ay 

y=0 


1 

Ax  j- 


=  2 


7rp 


2/3  dx  dy, 


=  ZE£  a4/t  =  _£.  Ma2,  where  if  is  the  mass  of  the  cone. 
10  10        ' 


MOMENTS   OF  INERTIA  359 

331.  Theorem.  The  moment  of  inertia  of  a  body  with  refer- 
ence to  any  line  is  equal  to  the  moment  of  inertia  of  the  body 
with  reference  to  a  parallel  line  through  the  center  of  gravity, 
plus  the  product  of  the  mass  of  the  body  and  the  square  of  the 
distance  between  the  two  lines. 

Proof.  Draw  a  plane  through,  the  given  line  and  the  center 
of  gravity  of  the  body.  Take  the  origin  at  the  center  of 
gravity,  the  x-axis  parallel  to  the  given  line,  and  the  y-axis  in 
the  plane  containing  the  a>axis  and  the  given  line. 

Let  Ix  denote  the  moment  of  inertia  of  the  body  with  refer- 
ence to  the  a>axis,  Ix  the  moment  of  inertia  of  the  body  with 
reference  to  the  given  line,  M  the  mass  of  the  body,  and  a  the 
distance  between  the  given  line  and  the  a>axis. 

Then       Ix  =  C(y2  +  z2)  dm,  and  I%  =  C\  (a  ±  yf  +  z2  \  dm. 

.-.  It— Ix=a2  I  dm  ±  2a  J  ydm, 

=  a2M  ±  2  a  \  y  dm. 

Now  |  y  dm  —  0,  since  the  origin  is  at  the  center  of  gravity 

of  the  body.     (See  Art.  319.) 

.-.  Il-Ix  =  a2M. 

.-.  It=Ix  +  a2M, 

which  establishes  the  theorem. 

J  r2dm 

332.  A  quantity  r0  can  always  be  found  such  that  r02=  — ^ — > 

where  M  is  the  mass  of  the  body,  and  I  i^dm  the  moment  of 
inertia   of   the  body  with  reference   to   some   point,  line,   or 

Plane-  fr'dm 

Definition.     This  quantity  r()  such  that  r2  =         —  is  called 

the  radius  of  gyration  of  the  body  with  reference  to  the  point, 
line,  or  plane, 


360       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

The  radius  of  gyration  determines  a  point  distant  from  a 
point,  line,  or  plane  such  that  if  the  whole  mass  of  the  body  be 
supposed  concentrated  at  that  point,  the  moment  of  inertia  of 
the  body  would  be  unchanged. 

EXERCISES 

Find  the  moment  of  inertia : 

1.  Of  a  parallelogram  of  base  b  and  altitude  h,  with  reference 
to  the  base;  with  reference  to  a  line  through  the  C.G.  parallel 
to  the  base.  A        r      57^  W 


Ans.  J6  =  — ;  Ig  = 


2.  Of  a  triangle  with  reference  to  the  base ;  with  reference 
to  a  line  through  the  CO.  parallel  to  the  base;  with  reference 
to  a  line  through  the  vertex  parallel  to  the  base. 

Ans.  Ib-  —  ,  I,-—,  J„_4 

3.  (a)  Of  a  square  with  reference  to  a  diagonal. 

(b)  Of  a  circle  with  reference  to  a  coordinate  axis;  with 
reference  to  the  origin. 

Ans.  (a)  I  =  j2'     (&)  J«  =  !f-5  Jo  =  !y- 

4.  Of  the  area  in  Exercise  23,  Chapter  XXXVI,  with  refer- 
ence to  the  base;  with  reference  to  a  line  through  the  C.G-. 
parallel  to  the  base. 

A        T       t,8l     ,9     +3N     T       t   5a4— 10 aH +  11  aH2-6at3+t4 
Ans.  Ib  =  -(a°+at-fi);  I,  =  - 2a_; 

5.  Prove  that  the  moment  of  inertia  of  a  system  of  areas  or 
solids  with  reference  to  any  axis  is  equal  to  the  sum  of  the 
moments  of  inertia  of  the  separate  areas  or  solids  with  reference 
to  that  axis. 

Find  the  moments  of  inertia  with  reference  to  the  horizontal 
and  vertical  gravity  axes  of  the  areas  described  in : 


MOMENTS   OF  INERTIA 


361 


6.  Exercise  24,  Chapter  XXXVI. 

Ans.  7",  =  3.963  bi.  in;  Iy  =  0.8597  bi.  in. 

7.  Exercise  25,  Chapter  XXXVI. 

Ans.  Ix  =  TV  bh*  -  j\ (b  -  t)(Ji  -2tf; 

Iy-z(h-2t)t  +  lstt>        4(A  +  26_20 

8.  Exercise  26,  Chapter  XXXVI. 

Ans.  Ix  =  2.298  bi.  in. ;  Iy  =  66.72  bi.  in. 

9.  Exercise  27,  Chapter  XXXVI. 

Ans.  J,  =  2.51  bi.  in. ;  L  =  24.65  bi.  in. 


K- 


10.  Determine  the  distance  d  so 
that  the  moment  of  inertia  of  the 
two  areas  of  the  figure  shall  be  the 
same  about  the  horizontal  and  verti- 
cal gravity  axes. 

Ans.  d  =  7.309".     Y 


4tt 


"T 


11.  Find  the  moment  of  inertia 
and  radius  of  gyration  of  the  hollow 
square  of  the  figure  with  reference 
to  a  line  through  the  C.G.  and  paral- 
lel to  a  side. 

Ans.  I=^\aA-(a-2ty\; 
k2  =  T\{a2+(a-2t)2l. 


12.  Find  the  moment  of  inertia  and  radius 
of  gyration  of  the  area  of  the  ring  of  the 
figure  with  reference  to  the  center;  with 
reference  to  a  diameter. 


Y a- h 


t 

1 

k 

to 



5 

1 

— - 

Ans.  /„  =  |[)-4-(r-J)4h  &2  =  ifr2+(V-02S 


Ia=\\^-{r-ty\;  W  =  l\r>+(r-tY\. 


362       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

13.   Find  the  radius  of  gyration  of  a  homogeneous  circular 
wire  with  reference  to  its  diameter.  .       ,         r 


V2 

14.  Find  the  moment  of  inertia  of  the  homogeneous  ellipsoid 

—  4-  y~  +  —  =  1,  with  reference  to  the  a?-axis. 

a      b      c  Ans.  I=^L{b2  +  c2). 

15.  Find  the  moment  of  inertia  of  a  homogeneous  sphere  of 
radius  a  with  reference  to  a  diameter. 

Ans.  1=  J-  Va2m,  where  m  is  the  mass  per  unit  of  volume. 

16.  Find  the  moment  of  inertia  and  radius  of  gyration  of  a 
homogeneous  right  circular  cylinder  of  length  I  and  radius  r, 
with  reference  to  its  axis :  with  reference  to  a  diameter  of  one 


end'  Ans.  Jasis  =  1MB2;  h  =  -^  •  Idiam.  =m(1  +  ^ 

17.  Show  that  if  either  coordinate  axis  is  an  axis  of  symme- 
try of  an  area  F,  then  J  |  xy  dx  dy  is  zero,  the  limits  being 
taken  to  embrace  the  whole  area. 

18.  If  Ia  is  the  moment  of  inertia  of  an  area  with  reference 
to  a  line  through  the  origin  making  an  angle  a  with  the  #-axis, 
and  if  either  coordinate  axis  is  an  axis  of  symmetry,  then 

Ia  =  cos2  a  Ix  +  sin2  a  Iy. 

19.  In  Exercise  18,  show  that  if  Ix  =  Ir  and  either  axis  is  an 
axis  of  symmetry,  then  Ia  is  the  same  for  all  values  of  a. 

20.  Prove  that  the  least  radius  of  gyration  of  the  figure  in 
Exercise  23,  Chapter  XXXVI,  with  reference  to  any  gravity 
axis  is  for  the  axis  sloped  45°  to  the  sides. 


INDEX 


[The  numbers  refer  to  the  pages.] 


Absolute  value  of  a  real  function,  5. 

Acceleration,  definitions  of,  283. 
determination  of,  283. 

Algebraic  function,  definition  of,  2. 
classification  of,  3. 

Algebraic  reduction  formulas,  181. 

Anti-logarithmic  functions,  defini- 
tions of,  3. 

Anti-trigonometric  functions,  defini- 
tions of,  3. 

Asymptote,  definition  of,  123. 

Attractive  force  varying  directly  as 
the  distance,  299. 

Attractive  force  varying  inversely 
as  the  square  of  the  distance, 
302. 

Center  of  gravity,  coordinates  of, 
348. 
definition  of,  344. 

Central  force,  321. 

Circumscribed  rectangles,  definitions 
of,  190. 

Concave  upwards,  concave  down- 
wards, definitions  of,  64. 

Concavity  upwards,  concavity 
downwards,  tests  for,  67. 

Constant,  definition  of,  1. 

Curvature,  definitions  of,  129. 
tests  for,  133. 

Curve  rising,  curve  falling,  defini- 
tions of,  61. 
tests  for,  63. 

Cycloidal  pendulum,  313. 

Density,  definition  of,  329. 
Dependent  variable,  definition  of,  2. 
Derivative,  definition  of,  31. 


Differentials,  definitions  of,  116. 
Displacement,  definition  of,  278. 
Double-valued  functions,  5. 
Dyne,  definition  of,  292. 

Elliptic  integrals,  definitions  of,  265. 

tables,  270. 
Evolute,  definition  of,  136. 

property  of,  139. 
Exponential     functions,     definition 
of,  3. 

Finite  discontinuity,  29. 
Force,  288. 

Function,    continually    decreasing, 
definition  of,  10. 
continually  increasing,  definition 

of,  10. 
continuous  or  discontinuous  for 

a  value  of  the  variable,  27. 
definition  of,  1. 
finite,  between  two  values  of  the 

variable,  27. 
infinite,  definition  of,  15,  26. 
infinite  negatively,  definition  of, 
16,  26. 
Functional  symbols,  5. 
Fundamental  forms  of  integration, 
154. 

Hyperbolic  functions,  85. 

Increment,  definition  of,  30. 
Indefinite    integral,    definition    of, 

153. 
Independent  variable,  definition  of, 

2. 
Infinite  discontinuity,  29. 
Infinitesimal,  definition  of,  20. 


363 


364 


INDEX 


Infinitesimals,  theorem  in,  211. 

Inflexion,  points  of,  67. 

Inscribed  rectangles,  definitions  of, 

190. 
Integral,  single,  double,  definitions 

of,  237. 
Integrand,  definition  of,  197. 
Integration  by  parts,  178.    J~ 
Integration  by  substitution,  169. 
Involute,  definition  of,  139. 

Kinetic  energy,  323. 

Law  of  the  mean,  98. 

Law  of  motion  of   a  moving  body, 

40. 
Limit,  definition  of,  12 
Limits,  lower,  upper,  definitions  of, 

213. 
Logarithmic    functions,     definition 

of,  3. 

Maclaurin's  theorem,  106. 
Mass,  definition  of,  289. 
Maximum  point,   minimum   point, 

definitions  of,  68. 
tests  for,  70. 
Maximum    or    minimum  value    of 

function,  74. 
Moment,  definition  of,  344. 
Moment  of    inertia,    definition    of, 

357. 
Momentum  or  quantity  of  motion, 

293. 
Motion  under  a  constant  force,  296. 
Multiple-valued  functions,  5. 

Osculating  circle,  135. 

Pappus'  theorems,  351. 

Partial   derivatives,   definitions    of, 

147. 
Periodic  time,  definition  of,  301. 
Planimeters,  276. 
Potential  energy,  325. 
Poundal,  definition  of,  292. 

Radius  of  curvature,  definition  of, 
135. 


Radius   of  gyration,    definition  of, 

359. 
Rational  fractions,  177. 
Rolle's  theorem,  97. 

Simple  harmonic  motion,  definition 

of,  302. 
Simple  pendulum,  313. 
Simpson's  rule,  274. 
Single-valued    functions,    definition 

of,  5. 
Slope  of  tangent  line,  polar  coordi- 
nates, 97. 
rectangular  coordinates,  35. 
Speed,  definitions  of,  38. 

units  of,  39. 
Subtangent,  subnormal,  polar  coor- 
dinates, 122. 
rectangular  coordinates,  121. 

Taylor's  theorem,  103. 

Trapezoidal  methods  of  approxima- 
tion, 271. 

Trigonometric  functions,  definition 
of,  3. 

Trigonometric  reduction  formulas, 
185. 

Units,  astronomical  system  of,  331. 

gravitational,  absolute,  291. 
Units  of  density,  329. 
Units  of  energy,  323. 
Units  of  rate  of  work,  322. 
Units  of  work,  318. 

Variable,  definition  of,  1. 

Variable  continually  decreasing, 
definition  of.  10. 

Variable  continually  increasing, 
definition  of,  10. 

Variable  infinite,  definition  of,  12. 
negatively,  definition  of,  17. 

Variables  involved  explicitly,  im- 
plicitly, 4. 

Velocity,  definition  of,  280. 

method  for  determination  of,  280. 

Weight,  definition  of,  290. 


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